Solution of Triangles for JEE: Every Formula, Proved and Solved

A triangle has six elements: three sides and three angles. Give me any three of them, one of which is a side, and the triangle is fixed. Everything else, every remaining side, every remaining angle, the area, the radius of the circle through the vertices, the radius of the circle inside it, is then a computation and nothing more.

This chapter is that computation, done properly. The sine rule and the cosine rule are the two engines. The half angle formulae are the bridge that turns any angle condition into a statement about $s$, $a$, $b$ and $c$. After that come the four centres, the four circles, and the handful of identities that JEE keeps coming back to. Everything below is proved, not asserted, and every worked example was re-derived from scratch before it was written down.

The six elements and what fixes a triangle

Write the standard labels once and never change them. In $\triangle ABC$ the angles at the vertices are $A$, $B$, $C$, and the side opposite each is $a$, $b$, $c$ respectively.

Triangle ABC with sides a, b, c opposite the angles A, B, C

Three facts hold in every triangle and are used constantly without comment:

$$A+B+C=180^\circ$$ $$a+b>c,\quad b+c>a,\quad c+a>b$$ $$a>b \iff A>B$$

The semi-perimeter is the single most useful abbreviation in the chapter:

$$s=\frac{a+b+c}{2}$$

so that $s-a=\dfrac{b+c-a}{2}$, and each of $s-a$, $s-b$, $s-c$ is strictly positive.

Example 1 (IITianForum). In $\triangle ABC$, $a=13$, $b=14$, $c=15$. Find $\cos A$, the area, the inradius and the circumradius.

▸ SHOW SOLUTION

Step 1. Write down $s$ and the area

$$s=\frac{13+14+15}{2}=21$$$$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$$$\Delta=\sqrt{21\cdot 8\cdot 7\cdot 6}$$$$\Delta=\sqrt{7056}=84$$

Step 2. Use the cosine rule for $\cos A$

$$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{196+225-169}{2\cdot 14\cdot 15}$$$$\cos A=\frac{252}{420}=\frac{3}{5}$$

Step 3. Read off the two radii

$$r=\frac{\Delta}{s}=\frac{84}{21}=4$$$$R=\frac{abc}{4\Delta}=\frac{13\cdot 14\cdot 15}{336}=\frac{65}{8}$$

Answer: $\cos A=\dfrac35$, $\Delta=84$, $r=4$, $R=\dfrac{65}{8}$.

This triangle is worth memorising. Its area, its inradius, its circumradius and all three exradii are rational, which makes it the cleanest possible test case for any identity you are unsure of.

The 13-14-15 triangle with its incircle and the values of s, area, r and R

Sine rule and the circumcircle

The circle through the three vertices is the circumcircle; its centre is the circumcentre $O$ and its radius is $R$.

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$

Proof. Draw the circumcircle and let $BA'$ be the diameter through $B$. The angle in a semicircle is a right angle, so $\angle BCA'=90^\circ$. Angles $\angle BA'C$ and $\angle BAC$ stand on the same arc $BC$, so $\angle BA'C=A$. In the right triangle $BCA'$,

$$a=BC=BA'\sin A=2R\sin A$$

The same argument at the other two vertices gives the other two ratios. If $A$ is obtuse the point $A'$ lies on the other arc and $\angle BA'C=180^\circ-A$, whose sine is still $\sin A$, so the result is unchanged.

Circumcircle with the diameter BA-prime used to prove a = 2R sin A

Three forms of the same statement are worth having in your hands at once: $$a=2R\sin A,\quad b=2R\sin B$$ $$c=2R\sin C$$ $$\sin A=\frac{a}{2R},\quad \sin B=\frac{b}{2R}$$ $$\sin C=\frac{c}{2R}$$ $$a:b:c=\sin A:\sin B:\sin C$$

The third line is the one that turns a condition on the sides into a condition on the angles, and back again. Whenever a question mixes sides and sines, convert everything to one or the other before doing anything else.

Example 2 (IITianForum). In $\triangle ABC$, $A=30^\circ$ and $a=8$. Find $R$, and find $b$ if $B=45^\circ$.

▸ SHOW SOLUTION

Step 1. Get $2R$ from the given pair

$$2R=\frac{a}{\sin A}=\frac{8}{\sin 30^\circ}=\frac{8}{1/2}=16$$

so $R=8$.

Step 2. Use the same constant for $b$

$$b=2R\sin B=16\sin 45^\circ=16\cdot\frac{\sqrt2}{2}$$$$b=8\sqrt2$$

Answer: $R=8$ and $b=8\sqrt2$.

Example 3 (JEE Main 2026, 4 April Shift 2). Let $A$ and $B$ be points on the two half-lines $x-\sqrt3\,|y|=\alpha$, $\alpha>0$, at a distance $\alpha$ from their point of intersection $P$. The segment $AB$ meets the angle bisector of the half-lines at $Q$. If $PQ=\dfrac92$ and $R$ is the circumradius of $\triangle PAB$, find $\dfrac{\alpha^2}{R}$.

▸ SHOW SOLUTION

The two half-lines meeting at P with the equilateral triangle PAB and the median PQ

Step 1. Split the modulus and find the apex angle

$x-\sqrt3\,|y|=\alpha$ is the pair of half-lines

$$x-\sqrt3\,y=\alpha \quad\text{and}\quad x+\sqrt3\,y=\alpha$$

meeting at $P(\alpha,0)$. Their slopes are $\dfrac{1}{\sqrt3}$ and $-\dfrac{1}{\sqrt3}$, so each makes $30^\circ$ with the $x$-axis and the angle at $P$ is $60^\circ$.

Step 2. Recognise the triangle

$PA=PB=\alpha$ with an included angle of $60^\circ$, so $\triangle PAB$ is equilateral of side $\alpha$.

Step 3. Use the median

The bisector of the apex angle is the $x$-axis, and it meets $AB$ at its midpoint $Q$. In an equilateral triangle the median is $\dfrac{\sqrt3}{2}$ times the side:

$$\frac{\sqrt3}{2}\,\alpha=\frac92 \implies \alpha=\frac{9}{\sqrt3}=3\sqrt3$$

Step 4. Circumradius of an equilateral triangle

$$R=\frac{a}{2\sin A}=\frac{\alpha}{2\sin 60^\circ}=\frac{\alpha}{\sqrt3}=3$$$$\frac{\alpha^2}{R}=\frac{27}{3}=9$$

Answer: $9$.

The whole question is the sine rule applied once, after one line of reading. The half-lines are a disguise for "isosceles with a $60^\circ$ apex", and that phrase always means equilateral.

Cosine rule

The cosine rule is what you reach for whenever an angle sits between two known sides, or whenever all three sides are known.

$$\cos A=\frac{b^2+c^2-a^2}{2bc}$$ $$\cos B=\frac{c^2+a^2-b^2}{2ca}$$ $$\cos C=\frac{a^2+b^2-c^2}{2ab}$$

Proof. Drop the perpendicular $AD$ from $A$ onto $BC$. In the right triangle $ABD$, $BD=c\cos B$ and $AD=c\sin B$, so $DC=a-c\cos B$. Pythagoras in $ADC$ gives

$$b^2=(c\sin B)^2+(a-c\cos B)^2$$

and using $\sin^2 B+\cos^2 B=1$ this collapses to

$$b^2=a^2+c^2-2ac\cos B$$

If $B$ is obtuse the foot $D$ falls outside $BC$ and $BD=c\cos B$ is negative, which is exactly what the algebra above already assumes. No separate case is needed.

Cosine rule proved by dropping one altitude and applying Pythagoras

Rearranged, the same statement reads $$a^2=b^2+c^2-2bc\cos A$$

and this is the form to use when the angle is given and the opposite side is wanted.

Example 4 (IITianForum). In $\triangle ABC$, $b=5$, $c=8$ and $A=60^\circ$. Find $a$ and the area.

▸ SHOW SOLUTION

Step 1. Cosine rule for the third side

$$a^2=25+64-2\cdot 5\cdot 8\cdot\cos 60^\circ$$$$a^2=89-80\cdot\frac12=89-40=49$$

so $a=7$.

Step 2. Area from two sides and the included angle

$$\Delta=\frac12 bc\sin A=\frac12\cdot 5\cdot 8\cdot\frac{\sqrt3}{2}$$$$\Delta=10\sqrt3$$

Answer: $a=7$ and $\Delta=10\sqrt3$.

Example 5 (JEE Main 2026, 21 January Shift 2). For a triangle $ABC$ let $\vec p=\overrightarrow{BC}$, $\vec q=\overrightarrow{CA}$, $\vec r=\overrightarrow{BA}$. If $|\vec p|=2\sqrt3$, $|\vec q|=2$ and $\cos\theta=\dfrac{1}{\sqrt3}$, where $\theta$ is the angle between $\vec p$ and $\vec q$, find $\big|\vec p\times(\vec q-3\vec r)\big|^2+3|\vec r|^2$.

▸ SHOW SOLUTION

Step 1. Find the relation between the three vectors

$$\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{BA}\implies \vec p+\vec q=\vec r$$

Step 2. Get $|\vec r|^2$ by the cosine rule

The angle between $\vec p$ and $\vec q$ as vectors is $\theta$, but the interior angle of the triangle at $C$ is $180^\circ-\theta$, because $\overrightarrow{CB}=-\vec p$. So

$$|\vec r|^2=|\vec p|^2+|\vec q|^2-2|\vec p||\vec q|\cos(180^\circ-\theta)$$$$|\vec r|^2=12+4+2(2\sqrt3)(2)\cdot\frac{1}{\sqrt3}$$$$|\vec r|^2=16+8=24$$

Step 3. Simplify the cross product before computing anything

$$\vec q-3\vec r=\vec q-3(\vec p+\vec q)=-3\vec p-2\vec q$$$$\vec p\times(-3\vec p-2\vec q)=-2(\vec p\times\vec q)$$

because $\vec p\times\vec p=\vec 0$.

Step 4. Take the square of the magnitude

$$\big|-2(\vec p\times\vec q)\big|^2=4|\vec p|^2|\vec q|^2\sin^2\theta$$$$\sin^2\theta=1-\frac13=\frac23$$$$=4\cdot 12\cdot 4\cdot\frac23=128$$

Step 5. Add the second term

$$128+3(24)=128+72=200$$

Answer: (4) $200$.

The vector notation is decoration. The only piece of real work is spotting that the interior angle at $C$ is $180^\circ-\theta$, not $\theta$; getting that wrong gives $|\vec r|^2=8$ and a wrong option that is on the paper.

Projection formulae

Every side is the sum of the projections of the other two onto it.

$$a=b\cos C+c\cos B$$ $$b=c\cos A+a\cos C$$ $$c=a\cos B+b\cos A$$

Proof. With $D$ the foot of the perpendicular from $A$ to $BC$, $BD=c\cos B$ and $DC=b\cos C$, and $a=BD+DC$. If $B$ is obtuse, $D$ lies outside the segment on the far side of $B$, $\cos B<0$, and $BD=c\cos B$ is a signed length, so the sum is still correct.

The projection formula in the acute case and the obtuse case

These three are the fastest route from a side condition to an angle condition, because substituting $a=2R\sin A$ turns the first into

$$\sin A=\sin B\cos C+\sin C\cos B$$ $$\sin A=\sin(B+C)$$

which is the statement $A+B+C=180^\circ$ in disguise. That is worth knowing: the projection formulae carry no information beyond the angle sum and the sine rule, so they never let you conclude something new on their own. What they do is convert an expression quickly.

Example 6 (IITianForum). In $\triangle ABC$, $a\cos A=b\cos B$. What can you say about the triangle?

▸ SHOW SOLUTION

Step 1. Convert sides to sines

Put $a=2R\sin A$ and $b=2R\sin B$:

$$2R\sin A\cos A=2R\sin B\cos B$$$$\sin 2A=\sin 2B$$

Step 2. Solve the sine equation, keeping both branches

$$2A=2B \quad\text{or}\quad 2A=180^\circ-2B$$$$A=B \quad\text{or}\quad A+B=90^\circ$$

Step 3. Read both branches as geometry

$A=B$ means the triangle is isosceles with $a=b$. And $A+B=90^\circ$ means $C=90^\circ$, so the triangle is right angled at $C$.

Answer: the triangle is isosceles ($a=b$) or right angled at $C$, and both genuinely occur.

The mark here is lost by cancelling. Writing $\sin 2A=\sin 2B\implies A=B$ throws away the second branch, and the option "isosceles or right angled" is always on the paper for exactly that reason.

Half angle formulae and the semi-perimeter

These are the bridge between angles and sides. Everything on the right is built from $s$, $a$, $b$, $c$ alone.

$$\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}$$ $$\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}$$ $$\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$$

Proof. Start from $\cos A=1-2\sin^2\dfrac{A}{2}$ and the cosine rule:

$$2\sin^2\frac{A}{2}=1-\frac{b^2+c^2-a^2}{2bc}$$ $$2\sin^2\frac{A}{2}=\frac{a^2-(b-c)^2}{2bc}$$ $$2\sin^2\frac{A}{2}=\frac{(a-b+c)(a+b-c)}{2bc}$$

Now $a-b+c=2(s-b)$ and $a+b-c=2(s-c)$, so

$$\sin^2\frac{A}{2}=\frac{(s-b)(s-c)}{bc}$$

Since $0<\dfrac{A}{2}<90^\circ$ the sine is positive and the square root is unambiguous. The cosine formula follows the same way from $\cos A=2\cos^2\dfrac{A}{2}-1$, and the tangent is their quotient.

The half angle is always acute, which is the whole reason these formulae are single valued while $\sin A=\dfrac{a}{2R}$ is not. That single fact resolves the ambiguous case later on.

Incircle touching the three sides, with tangent lengths s-a, s-b and s-c

The picture explains where $s-a$ comes from. The two tangents to the incircle from $A$ are equal; call that length $x$, and similarly $y$ from $B$ and $z$ from $C$. Then

$$y+z=a,\quad z+x=b,\quad x+y=c$$ $$x+y+z=s$$ $$x=s-a,\quad y=s-b,\quad z=s-c$$

and from the right triangle formed by $A$, the incentre and a point of contact,

$$\tan\frac{A}{2}=\frac{r}{s-a}$$

Example 7 (IITianForum). For the triangle with $a=13$, $b=14$, $c=15$, find $\tan\dfrac{A}{2}$ and confirm it against $r$ and $s-a$.

▸ SHOW SOLUTION

Step 1. Use the half angle formula

With $s=21$: $s-a=8$, $s-b=7$, $s-c=6$.

$$\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\sqrt{\frac{7\cdot 6}{21\cdot 8}}$$$$\tan\frac{A}{2}=\sqrt{\frac{42}{168}}=\sqrt{\frac14}=\frac12$$

Step 2. Cross-check with the tangent length

From Example 1, $r=4$, and $s-a=8$:

$$\frac{r}{s-a}=\frac{4}{8}=\frac12$$

The two agree.

Step 3. Sanity check against $\cos A$

$\tan\dfrac{A}{2}=\dfrac12$ gives $\cos A=\dfrac{1-\tan^2(A/2)}{1+\tan^2(A/2)}=\dfrac{1-\frac14}{1+\frac14}=\dfrac35$, which is what Example 1 found.

Answer: $\tan\dfrac{A}{2}=\dfrac12$.

Napier's analogy and Mollweide's check

Napier's analogy converts a ratio of sides into a difference of angles, which is exactly what you need when two sides and the included angle are given.

$$\tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot\frac{C}{2}$$ $$\tan\frac{B-C}{2}=\frac{b-c}{b+c}\cot\frac{A}{2}$$ $$\tan\frac{C-A}{2}=\frac{c-a}{c+a}\cot\frac{B}{2}$$

Proof. By the sine rule,

$$\frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A+\sin B}$$ $$=\frac{2\cos\frac{A+B}{2}\sin\frac{A-B}{2}}{2\sin\frac{A+B}{2}\cos\frac{A-B}{2}}$$ $$=\tan\frac{A-B}{2}\cot\frac{A+B}{2}$$

and $\dfrac{A+B}{2}=90^\circ-\dfrac{C}{2}$, so $\cot\dfrac{A+B}{2}=\tan\dfrac{C}{2}$. Rearranging gives the result.

Mollweide's formulae are the companion pair, and they are the best available check on a finished solution because every one of the six elements appears in each:

$$\frac{a+b}{c}=\frac{\cos\frac{A-B}{2}}{\sin\frac{C}{2}}$$ $$\frac{a-b}{c}=\frac{\sin\frac{A-B}{2}}{\cos\frac{C}{2}}$$

Example 8 (IITianForum). In $\triangle ABC$, $a=6$, $b=3$ and $C=60^\circ$. Solve the triangle.

▸ SHOW SOLUTION

The triangle a=6, b=3, C=60 degrees solved by Napier's analogy

Step 1. Napier for the difference of the other two angles

$$\tan\frac{A-B}{2}=\frac{6-3}{6+3}\cot 30^\circ=\frac{3}{9}\cdot\sqrt3$$$$\tan\frac{A-B}{2}=\frac{\sqrt3}{3}=\tan 30^\circ$$

so $\dfrac{A-B}{2}=30^\circ$ and $A-B=60^\circ$.

Step 2. Combine with the angle sum

$$A+B=180^\circ-60^\circ=120^\circ$$$$A=90^\circ,\qquad B=30^\circ$$

Step 3. Third side by the cosine rule

$$c^2=36+9-2\cdot 6\cdot 3\cdot\frac12=45-18=27$$$$c=3\sqrt3$$

Step 4. Check with Mollweide

$$\frac{a-b}{c}=\frac{3}{3\sqrt3}=\frac{1}{\sqrt3}$$$$\frac{\sin\frac{A-B}{2}}{\cos\frac{C}{2}}=\frac{\sin 30^\circ}{\cos 30^\circ}=\frac{1/2}{\sqrt3/2}=\frac{1}{\sqrt3}$$

The two agree, so all six elements are consistent.

Answer: $A=90^\circ$, $B=30^\circ$, $c=3\sqrt3$.

Area of a triangle: seven ways

Each of these is the same number $\Delta$, written in terms of whatever you happen to know.

Area from a base and the height b sin C

Proof of the first. The height from $A$ onto $BC$ is $b\sin C$, so $\Delta=\frac12 a\cdot b\sin C$. Proof of Heron's. Square the first form and use $\sin^2 C=(1-\cos C)(1+\cos C)$ with the cosine rule; the four factors that appear are exactly $2s$, $2(s-a)$, $2(s-b)$, $2(s-c)$ divided by $4ab$. Proof of the third. Substitute $\sin C=\dfrac{c}{2R}$ into the first. Proof of the fourth. Join the incentre to the three vertices, splitting the triangle into three triangles of heights $r$ and bases $a$, $b$, $c$, so $\Delta=\frac12 r(a+b+c)=rs$.

The three altitudes $p_1$, $p_2$, $p_3$ from $A$, $B$, $C$ satisfy $p_1=\dfrac{2\Delta}{a}$ and so on, from which

$$\frac{1}{p_1}+\frac{1}{p_2}+\frac{1}{p_3}=\frac{a+b+c}{2\Delta}=\frac1r$$ $$p_1p_2p_3=\frac{8\Delta^3}{abc}=\frac{(abc)^2}{8R^3}$$

Example 9 (IITianForum). Find the three altitudes of the triangle with sides $13$, $14$, $15$, and verify $\dfrac{1}{p_1}+\dfrac{1}{p_2}+\dfrac{1}{p_3}=\dfrac1r$.

▸ SHOW SOLUTION

Step 1. Use $\Delta=84$ from Example 1

$$p_1=\frac{2\Delta}{a}=\frac{168}{13}$$$$p_2=\frac{2\Delta}{b}=\frac{168}{14}=12$$$$p_3=\frac{2\Delta}{c}=\frac{168}{15}=\frac{56}{5}$$

Step 2. Add the reciprocals over a common denominator

$$\frac{13}{168}+\frac{1}{12}+\frac{5}{56}=\frac{13}{168}+\frac{14}{168}+\frac{15}{168}$$$$=\frac{42}{168}=\frac14$$

Step 3. Compare with $\dfrac1r$

$r=4$, so $\dfrac1r=\dfrac14$. They match.

Answer: $\dfrac{168}{13}$, $12$, $\dfrac{56}{5}$, and the identity holds.

Notice why the identity is obvious once written this way: the reciprocals are $\dfrac{a}{2\Delta}$, $\dfrac{b}{2\Delta}$, $\dfrac{c}{2\Delta}$, which sum to $\dfrac{2s}{2\Delta}=\dfrac{s}{\Delta}=\dfrac1r$.

Medians, Apollonius and the m-n theorem

The median from $A$ meets $BC$ at its midpoint $D$. Its length is

$$m_a=\frac12\sqrt{2b^2+2c^2-a^2}$$

Proof (Apollonius). Apply the cosine rule in $\triangle ABD$ and in $\triangle ACD$ at the two supplementary angles $\angle ADB$ and $\angle ADC$. The cosine terms are equal and opposite, so adding the two equations kills them:

$$b^2+c^2=2m_a^2+\frac{a^2}{2}$$

which rearranges to the formula above. In words: the sum of the squares of two sides is twice the square of the median to the third side plus twice the square of half that side.

Median AD of a triangle, with Apollonius theorem stated

Two consequences worth carrying:

$$m_a^2+m_b^2+m_c^2=\frac34\left(a^2+b^2+c^2\right)$$ $$GA^2+GB^2+GC^2=\frac{a^2+b^2+c^2}{3}$$

where $G$ is the centroid. The second follows from the first because $GA=\frac23 m_a$.

For a general cevian, not just a median, the tool is the $m$-$n$ theorem. If $D$ lies on $BC$ with $BD:DC=m:n$, and $\theta=\angle ADC$, then

$$(m+n)\cot\theta=n\cot B-m\cot C$$ $$(m+n)\cot\theta=m\cot\alpha-n\cot\beta$$

where $\alpha=\angle BAD$ and $\beta=\angle DAC$.

The m-n theorem for a cevian dividing BC in the ratio m to n

Proof. Write the area of $\triangle ABD$ two ways and the area of $\triangle ADC$ two ways, or apply the sine rule in each of the two sub-triangles and divide. Either route gives $\dfrac{BD}{DC}=\dfrac{\sin\alpha}{\sin\beta}\cdot\dfrac{\sin C}{\sin B}$, and expanding $\theta=B+\alpha$ turns that into the cotangent form.

Example 10 (IITianForum). In $\triangle ABC$, $B=30^\circ$ and $C=45^\circ$. If $AD$ is the median to $BC$, find $\cot(\angle ADC)$.

▸ SHOW SOLUTION

Step 1. A median is the case $m=n$

$$(m+n)\cot\theta=n\cot B-m\cot C$$

with $m=n$ becomes

$$2\cot\theta=\cot B-\cot C$$

Step 2. Substitute the two angles

$$2\cot\theta=\cot 30^\circ-\cot 45^\circ=\sqrt3-1$$$$\cot\theta=\frac{\sqrt3-1}{2}$$

Answer: $\cot(\angle ADC)=\dfrac{\sqrt3-1}{2}$.

Worth noting: $\cot\theta>0$, so $\theta$ is acute, which is the expected picture when $C>B$. If you get a negative cotangent for $\angle ADC$ in a problem like this, check which of the two angles at $D$ you have actually computed.

A median with base angles 30 and 45 degrees

Example 11 (JEE Main 2025, 4 April Shift 2). The three sides of a triangle $ABC$ are given by the vectors $2\hat i-\hat j+\hat k$, $\hat i-3\hat j-5\hat k$ and $3\hat i-4\hat j-4\hat k$. If $G$ is the centroid, find $6\left(|\overrightarrow{AG}|^2+|\overrightarrow{BG}|^2+|\overrightarrow{CG}|^2\right)$.

▸ SHOW SOLUTION

Step 1. Recognise the identity that does all the work

$$GA^2+GB^2+GC^2=\frac{a^2+b^2+c^2}{3}$$

so the answer is $2\left(a^2+b^2+c^2\right)$, and the three side vectors are given, so nothing needs to be located.

Step 2. Square the three side lengths

$$a^2+b^2+c^2=6+35+41$$$$a^2+b^2+c^2=82$$

Step 3. Multiply

$$6\left(GA^2+GB^2+GC^2\right)=2\cdot 82=164$$

Answer: $164$.

A quick consistency check that the three vectors really do close into a triangle: $(2,-1,1)+(1,-3,-5)=(3,-4,-4)$, so with suitable orientations they sum to zero. If they had not, the question would have no triangle at all.

The four centres

The circumcentre, incentre, centroid and orthocentre of the same triangle

The coordinate forms are worth having exactly right, because JEE asks for them directly:

$$G=\frac{A+B+C}{3}$$ $$I=\frac{aA+bB+cC}{a+b+c}$$

The incentre is the side-weighted average, with the weight $a$ attached to the vertex $A$ opposite it. The centroid is the plain average. Confusing the two is the commonest slip in this part of the chapter.

Distances from a vertex, all of which follow from the sine rule in the small triangles the centres create:

$$OA=OB=OC=R$$ $$IA=r\,\operatorname{cosec}\frac{A}{2}=4R\sin\frac{B}{2}\sin\frac{C}{2}$$ $$HA=2R\cos A$$

and the distance from $H$ to the side $BC$ is $2R\cos B\cos C$.

Example 12 (JEE Main 2026, 2 April Shift 1). The midpoints of the sides of a triangle $ABC$ are $\left(\dfrac52,7\right)$, $\left(\dfrac52,3\right)$ and $(4,5)$. If its incentre is $(h,k)$, find $3h+k$.

▸ SHOW SOLUTION

The medial triangle, the recovered vertices and the incentre

Step 1. Recover the vertices

If $M_A$, $M_B$, $M_C$ are the midpoints of $BC$, $CA$, $AB$, then

$$A=M_B+M_C-M_A$$

and cyclically. Taking $M_A=\left(\frac52,7\right)$, $M_B=\left(\frac52,3\right)$, $M_C=(4,5)$:

$$A=(4,1),\quad B=(4,9),\quad C=(1,5)$$

Step 2. Find the side lengths

$$a=BC=\sqrt{9+16}=5$$$$b=CA=\sqrt{9+16}=5$$$$c=AB=\sqrt{0+64}=8$$

Step 3. Weight the vertices correctly

$$h=\frac{5(4)+5(4)+8(1)}{18}=\frac{48}{18}=\frac{8}{3}$$$$k=\frac{5(1)+5(9)+8(5)}{18}=\frac{90}{18}=5$$

Step 4. Combine

$$3h+k=3\cdot\frac83+5=8+5=13$$

Answer: (3) $13$.

The weights are $a$ on $A$, $b$ on $B$, $c$ on $C$. Here the side opposite $A(4,1)$ is $BC=5$, not the side $AB=8$ that touches it. Attach each weight to the vertex it faces, never to the side it lies on.

Example 13 (JEE Main 2025, 22 January Shift 2). Let $A(6,8)$, $B(10\cos\alpha,-10\sin\alpha)$ and $C(-10\sin\alpha,10\cos\alpha)$ be the vertices of a triangle. If $L(a,9)$ is its orthocentre and $G(h,k)$ its centroid, find $5a-3h+6k+100\sin 2\alpha$.

▸ SHOW SOLUTION

Step 1. Notice where the vertices lie

$$6^2+8^2=100$$$$100\cos^2\alpha+100\sin^2\alpha=100$$

All three vertices satisfy $x^2+y^2=100$, so the circumcentre is the origin.

Step 2. Use $H=A+B+C$

$$L_y=8-10\sin\alpha+10\cos\alpha=9$$$$10(\cos\alpha-\sin\alpha)=1$$$$L_x=a=6+10\cos\alpha-10\sin\alpha$$$$a=6+1=7$$

Step 3. The centroid is one third of that

$$h=\frac73,\qquad k=\frac93=3$$

Step 4. Square the condition to reach $\sin 2\alpha$

$$(\cos\alpha-\sin\alpha)^2=\frac{1}{100}$$$$1-\sin 2\alpha=\frac{1}{100}\implies 100\sin 2\alpha=99$$

Step 5. Assemble

$$5(7)-3\left(\frac73\right)+6(3)+99$$$$=35-7+18+99=145$$

Answer: $145$.

Example 14 (JEE Main 2026, 22 January Shift 2). Statement (S1): if $A(5,-1)$ and $B(-2,3)$ are two vertices of a triangle whose orthocentre is $(0,0)$, then the third vertex is $(-4,-7)$. Is (S1) correct?

▸ SHOW SOLUTION

Step 1. Write the two perpendicularity conditions

Let $C(h,k)$ and $O(0,0)$ be the orthocentre. Then $AO\perp BC$ and $BO\perp AC$.

Step 2. First condition

$$m_{AO}=\frac{0-(-1)}{0-5}=-\frac15$$$$m_{BC}=\frac{k-3}{h+2}$$$$-\frac15\cdot\frac{k-3}{h+2}=-1\implies k=5h+13$$

Step 3. Second condition

$$m_{BO}=\frac{0-3}{0+2}=-\frac32,\qquad m_{AC}=\frac{k+1}{h-5}$$$$-\frac32\cdot\frac{k+1}{h-5}=-1\implies 2h-3k=13$$

Step 4. Solve the pair

$$2h-3(5h+13)=13\implies -13h=52$$$$h=-4$$$$k=5(-4)+13=-7$$

Answer: (S1) is correct: the third vertex is $(-4,-7)$.

Use the dot product form $\overrightarrow{OA}\cdot\overrightarrow{BC}=0$ rather than slopes when either line risks being vertical. Here neither is, but the habit costs nothing.

Example 15 (JEE Main 2026, 4 April Shift 1). The vertex $A$ of triangle $ABC$ is $(1,2)$, the midpoint of $AB$ is $(5,-1)$, and the centroid is $(3,4)$. If the circumcentre is $(\alpha,\beta)$, find $21(\alpha+\beta)$.

▸ SHOW SOLUTION

Circumcentre found from a vertex, a midpoint and the centroid

Step 1. Get $B$ from the midpoint

$$B=2(5,-1)-(1,2)=(9,-4)$$

Step 2. Get $C$ from the centroid

$$C=3(3,4)-(1,2)-(9,-4)=(-1,14)$$

Step 3. Impose equal distances from the circumcentre

$PA^2=PB^2$ gives

$$4\alpha-3\beta=23$$

and $PA^2=PC^2$ gives

$$-\alpha+6\beta=48$$

Step 4. Solve and combine

$$\alpha=6\beta-48\implies 21\beta=215$$$$\beta=\frac{215}{21}$$$$\alpha=\frac{282}{21}$$$$21(\alpha+\beta)=282+215=497$$

Answer: (3) $497$.

The question hands you $21(\alpha+\beta)$ rather than $\alpha+\beta$ precisely because the denominator is $21$. When a question multiplies the answer by an odd-looking integer, that integer is usually the denominator you are about to produce, and it is a free check that you are on the right line.

Incircle, excircles and the identities they satisfy

The incircle touches all three sides from the inside; its radius is $r$ and its centre is $I$. There are three escribed circles, or excircles: the one opposite $A$ touches the side $BC$ and the other two sides produced, and its radius is $r_1$. Similarly $r_2$ opposite $B$ and $r_3$ opposite $C$.

Escribed circle opposite A, touching BC and the two produced sides

The inradius, in the forms you will actually use:

$$r=\frac{\Delta}{s}$$ $$r=(s-a)\tan\frac{A}{2}$$ $$r=(s-b)\tan\frac{B}{2}=(s-c)\tan\frac{C}{2}$$ $$r=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$ $$r=\frac{a\sin\frac{B}{2}\sin\frac{C}{2}}{\cos\frac{A}{2}}$$

The exradii:

$$r_1=\frac{\Delta}{s-a},\quad r_2=\frac{\Delta}{s-b},\quad r_3=\frac{\Delta}{s-c}$$ $$r_1=s\tan\frac{A}{2}$$ $$r_1=4R\sin\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$$

The last one has a clean pattern: $r$ is $4R$ times three sines; each $r_i$ swaps two of those sines for cosines, keeping the sine at the vertex it is opposite. Notice also that $r_1=s\tan\frac{A}{2}$ against $r=(s-a)\tan\frac{A}{2}$, so

$$\frac{r_1}{r}=\frac{s}{s-a}$$

Six identities that JEE returns to again and again, all of them provable in a line from the forms above:

$$r_1+r_2+r_3-r=4R$$ $$\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac1r$$ $$r_1r_2+r_2r_3+r_3r_1=s^2$$ $$r_1r_2r_3=rs^2=\frac{\Delta^3}{(s-a)(s-b)(s-c)}$$ $$\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}=\frac{1}{2Rr}$$ $$\frac{1}{r^2}+\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}=\frac{a^2+b^2+c^2}{\Delta^2}$$

Proof of the second. $\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3}=\dfrac{(s-a)+(s-b)+(s-c)}{\Delta}=\dfrac{3s-2s}{\Delta}=\dfrac{s}{\Delta}=\dfrac1r$. Every other identity in that list falls the same way: write each radius as $\Delta$ over a linear expression in $s$, and let the algebra collapse.

Incircle and the three excircles of the 12-16-20 triangle

Example 16 (IITianForum). In a triangle the three exradii are $r_1=8$, $r_2=12$, $r_3=24$. Find $r$, $R$, $s$, the area and the three sides.

▸ SHOW SOLUTION

Step 1. Get $r$ from the reciprocal identity

$$\frac1r=\frac18+\frac{1}{12}+\frac{1}{24}=\frac{3+2+1}{24}=\frac{6}{24}$$

so $r=4$.

Step 2. Get $R$ from the sum identity

$$4R=r_1+r_2+r_3-r$$$$4R=8+12+24-4=40$$

so $R=10$.

Step 3. Get $s$ from the pairwise product identity

$$s^2=r_1r_2+r_2r_3+r_3r_1$$$$s^2=96+288+192=576$$

so $s=24$, and $\Delta=rs=4\cdot 24=96$.

Step 4. Recover the sides

$$s-a=\frac{\Delta}{r_1}=\frac{96}{8}=12\implies a=12$$$$s-b=\frac{96}{12}=8\implies b=16$$$$s-c=\frac{96}{24}=4\implies c=20$$

Step 5. Check

$12$, $16$, $20$ is $4$ times $3,4,5$, so the triangle is right angled at $C$. Then $\Delta=\frac12\cdot 12\cdot 16=96$ and $R=\dfrac{c}{2}=10$, both of which agree.

Answer: $r=4$, $R=10$, $s=24$, $\Delta=96$, sides $12$, $16$, $20$.

Example 17 (IITianForum). Prove that in any triangle $\cos A+\cos B+\cos C=1+\dfrac{r}{R}$.

▸ SHOW SOLUTION

Step 1. Convert the left side to half angles

$$\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$$$$\cos C=1-2\sin^2\frac{C}{2}$$

and $\cos\dfrac{A+B}{2}=\sin\dfrac{C}{2}$, so the sum is

$$1+2\sin\frac{C}{2}\left(\cos\frac{A-B}{2}-\sin\frac{C}{2}\right)$$

Step 2. Replace the second $\sin\frac{C}{2}$

$\sin\dfrac{C}{2}=\cos\dfrac{A+B}{2}$, and

$$\cos\frac{A-B}{2}-\cos\frac{A+B}{2}=2\sin\frac{A}{2}\sin\frac{B}{2}$$

Step 3. Collect

$$\cos A+\cos B+\cos C$$$$=1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$

Step 4. Recognise the last term

$r=4R\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}$, so the bracket is exactly $\dfrac{r}{R}$.

$$\cos A+\cos B+\cos C=1+\frac{r}{R}$$

Since $r>0$ and $R>0$, this proves the standard bound $\cos A+\cos B+\cos C>1$ at once, and with $R\ge 2r$ from the next section it gives the upper bound $\dfrac32$.

Euler line and the distances between centres

Circumcentre, centroid, nine-point centre and orthocentre on the Euler line

In every non-equilateral triangle the circumcentre $O$, the centroid $G$ and the orthocentre $H$ are collinear, and

$$OG:GH=1:2$$

Proof. With $O$ at the origin, define $P=A+B+C$. Then $\overrightarrow{AP}=B+C$, which is perpendicular to $\overrightarrow{BC}=C-B$ because $(B+C)\cdot(C-B)=|C|^2-|B|^2=R^2-R^2=0$. So $P$ lies on the altitude from $A$, and by symmetry on all three: $P=H$. Since $G=\dfrac{A+B+C}{3}=\dfrac{H}{3}$, the three points $O$, $G$, $H$ are collinear with $OG:GH=1:2$.

The midpoint $N$ of $OH$ is the centre of the nine-point circle, which passes through the three side midpoints, the three feet of the altitudes and the three midpoints of $AH$, $BH$, $CH$. Its radius is $\dfrac{R}{2}$.

The three standard distances:

$$OI^2=R^2-2Rr$$ $$OH^2=R^2\left(1-8\cos A\cos B\cos C\right)$$ $$II_1=4R\sin\frac{A}{2}$$

Proof of the first. In $\triangle OAI$, $OA=R$ and $AI=4R\sin\frac{B}{2}\sin\frac{C}{2}$, and the angle $\angle OAI$ equals $\dfrac{|B-C|}{2}$ because $\angle OAB=90^\circ-C$ while $\angle IAB=\dfrac{A}{2}$. The cosine rule then gives, after using $r=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$,

$$OI^2=R^2-8R^2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$ $$OI^2=R^2-2Rr$$

Example 18 (JEE Main 2025, 24 January Shift 2). The position vectors of the three vertices of a triangle are $4\vec p+\vec q-3\vec r$, $-5\vec p+\vec q+2\vec r$ and $2\vec p-\vec q+2\vec r$. If the orthocentre is $\dfrac{\vec p+\vec q+\vec r}{4}$ and the circumcentre is $\alpha\vec p+\beta\vec q+\gamma\vec r$, find $\alpha+2\beta+5\gamma$.

▸ SHOW SOLUTION

Step 1. Compute the centroid, which needs no geometry

The coefficients of $\vec p$, $\vec q$, $\vec r$ in the sum of the three position vectors are $4-5+2=1$, $1+1-1=1$ and $-3+2+2=1$, so

$$3G=\vec p+\vec q+\vec r$$$$G=\frac{\vec p+\vec q+\vec r}{3}$$

Step 2. Use the Euler line ratio

$G$ divides $HO$ (from orthocentre to circumcentre) in the ratio $2:1$, so

$$3G=2O+H$$$$2O=3G-H=(\vec p+\vec q+\vec r)-\frac{\vec p+\vec q+\vec r}{4}$$$$2O=\frac34(\vec p+\vec q+\vec r)$$

Step 3. Read off the coefficients

$$O=\frac38(\vec p+\vec q+\vec r)$$

so $\alpha=\beta=\gamma=\dfrac38$.

Step 4. Combine

$$\alpha+2\beta+5\gamma=\frac38(1+2+5)=\frac{24}{8}=3$$

Answer: (1) $3$.

Nothing about $\vec p$, $\vec q$, $\vec r$ was needed beyond their being a basis, which is why the centroid could be written down immediately. The Euler line ratio $OG:GH=1:2$ is the entire question.

Example 19 (JEE Main 2025, 23 January Shift 1). The area of $\triangle PQR$ with $P(5,4)$, $Q(-2,4)$ and $R(a,b)$ is $35$. Its orthocentre is $\left(2,\dfrac{14}{5}\right)$ and its centroid is $(c,d)$. Find $c+2d$.

▸ SHOW SOLUTION

Step 1. Use the area to pin $b$

$PQ$ lies along $y=4$ and has length $7$, so

$$\frac12\cdot 7\cdot|b-4|=35\implies |b-4|=10$$

giving $b=14$ or $b=-6$.

Step 2. Impose the orthocentre conditions

With $O\left(2,\frac{14}{5}\right)$, require $\overrightarrow{OP}\cdot\overrightarrow{QR}=0$ and $\overrightarrow{OQ}\cdot\overrightarrow{PR}=0$. Solving the pair gives the single point

$$R=(2,-6)$$

which is consistent with $b=-6$ from Step 1. The value $b=14$ does not satisfy both perpendicularity conditions.

Step 3. Centroid

$$c=\frac{5-2+2}{3}=\frac53$$$$d=\frac{4+4-6}{3}=\frac23$$

Step 4. Combine

$$c+2d=\frac53+\frac43=\frac93=3$$

Answer: (2) $3$.

Example 20 (IITianForum). Show that a triangle with $R=2r$ must be equilateral.

▸ SHOW SOLUTION

Step 1. Use the distance formula for $OI$

$$OI^2=R^2-2Rr$$

Step 2. Substitute the condition

$R=2r$ gives $2Rr=R^2$, so

$$OI^2=R^2-R^2=0$$

Step 3. Interpret

$OI=0$ means the circumcentre and the incentre coincide. If $O=I$ then every perpendicular bisector is also an angle bisector, so each angle bisector meets the opposite side at its midpoint, which forces the two sides at that vertex to be equal. Applying this at all three vertices gives $a=b=c$.

Answer: the triangle is equilateral.

The converse is immediate: in an equilateral triangle $R=\dfrac{a}{\sqrt3}$ and $r=\dfrac{a}{2\sqrt3}$, so $R=2r$.

Pedal and excentral triangles

The pedal triangle $DEF$ has as its vertices the feet of the three altitudes. Pedal triangle DEF formed by the feet of the three altitudes For an acute triangle:

$$\angle FDE=180^\circ-2A$$ $$EF=a\cos A,\quad FD=b\cos B$$ $$DE=c\cos C$$ $$\text{area}=\frac{R^2}{2}\sin 2A\sin 2B\sin 2C$$ $$\text{circumradius}=\frac{R}{2}$$ $$\text{inradius}=2R\cos A\cos B\cos C$$

The circumradius $\dfrac{R}{2}$ is no accident: the pedal triangle is inscribed in the nine-point circle. If the original triangle is obtuse at $A$, the angles of the pedal triangle become $2A-180^\circ$, $2B$, $2C$ and the side $EF$ becomes $-a\cos A$, which is positive because $\cos A<0$.

The excentral triangle $I_1I_2I_3$ has the three excentres as vertices.

Excentral triangle with the incentre as its orthocentre

$$\angle I_2I_1I_3=90^\circ-\frac{A}{2}$$ $$I_2I_3=4R\cos\frac{A}{2}$$ $$\text{circumradius}=2R$$ $$\text{area}=8R^2\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$$

and $I$ is the orthocentre of $I_1I_2I_3$, while the original triangle $ABC$ is its pedal triangle. That single sentence generates every formula in the list: apply the pedal results to $I_1I_2I_3$ and you recover the ones for $ABC$.

Example 21 (IITianForum). In an acute triangle with $R=5$ and angles $45^\circ$, $60^\circ$, $75^\circ$, find the sides of the pedal triangle and its circumradius.

▸ SHOW SOLUTION

Step 1. Find the sides of the original triangle

$$a=2R\sin 45^\circ=10\cdot\frac{1}{\sqrt2}=5\sqrt2$$$$b=2R\sin 60^\circ=10\cdot\frac{\sqrt3}{2}=5\sqrt3$$$$c=2R\sin 75^\circ$$

Step 2. Apply $EF=a\cos A$ and its partners

$$EF=a\cos A=5\sqrt2\cdot\frac{1}{\sqrt2}=5$$$$FD=b\cos B=5\sqrt3\cdot\frac12=\frac{5\sqrt3}{2}$$$$DE=c\cos C=10\sin 75^\circ\cos 75^\circ$$$$DE=5\sin 150^\circ$$$$DE=\frac52$$

Step 3. Circumradius of the pedal triangle

$$\frac{R}{2}=\frac52$$

Answer: sides $5$, $\dfrac{5\sqrt3}{2}$, $\dfrac52$, circumradius $\dfrac52$.

The pattern $a\cos A=2R\sin A\cos A=R\sin 2A$ is the quickest form of all: the pedal sides are $R\sin 2A$, $R\sin 2B$, $R\sin 2C$, which is a circle of radius $\dfrac{R}{2}$ read through the sine rule.

Solving a triangle: four cases and the ambiguous one

Three sides given (SSS). Use $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, or $\tan\dfrac{A}{2}$ if you also want the area or the radii. One triangle, always, provided the three triangle inequalities hold.

Two sides and the included angle (SAS). Use the cosine rule for the third side, then the sine rule or Napier's analogy for the remaining angles. One triangle, always.

Two angles and one side (AAS or ASA). Get the third angle from the angle sum, then two applications of the sine rule. One triangle, always, provided the two given angles sum to less than $180^\circ$.

Two sides and a non-included angle (SSA). This is the ambiguous case. Given $b$, $c$ and $B$, the sine rule gives

$$\sin C=\frac{c\sin B}{b}$$

and $\sin C$ does not determine $C$, because $C$ and $180^\circ-C$ have the same sine. Four things can happen.

The four sub-cases of the ambiguous SSA situation

Everything in the table is the statement $\sin C=\dfrac{c\sin B}{b}$ read carefully. If the right side exceeds $1$ there is no $C$; if it equals $1$ there is one, namely $90^\circ$; if it is less than $1$ there are two candidates and you must test whether each gives a positive third angle. When $b\ge c$ we have $B\ge C$, so $C$ cannot be the obtuse one and only the acute root survives.

Example 22 (IITianForum). In $\triangle ABC$, $b=6$, $c=10$ and $B=30^\circ$. How many triangles are possible, and what are the values of $a$?

▸ SHOW SOLUTION

The arc of radius 6 cutting the ray from B in two points

Step 1. Test the condition

$$c\sin B=10\sin 30^\circ=5$$

Since $5<6<10$, that is $c\sin B<b<c$, there are two triangles.

Step 2. Use the cosine rule as a quadratic in $a$

$$36=a^2+100-2\cdot a\cdot 10\cos 30^\circ$$$$a^2-10\sqrt3\,a+64=0$$

Step 3. Solve

$$a=\frac{10\sqrt3\pm\sqrt{300-256}}{2}$$$$a=\frac{10\sqrt3\pm\sqrt{44}}{2}$$$$a=5\sqrt3\pm\sqrt{11}$$

Both roots are positive, which confirms two triangles.

Step 4. Read the two shapes

$$a\approx 11.98\quad\text{and}\quad a\approx 5.34$$

The larger goes with the acute $C$, the smaller with the obtuse one.

Answer: two triangles, with $a=5\sqrt3+\sqrt{11}$ and $a=5\sqrt3-\sqrt{11}$.

Equilateral and isosceles shortcuts

For an equilateral triangle of side $a$, every quantity in this chapter collapses:

$$\Delta=\frac{\sqrt3}{4}a^2,\qquad R=\frac{a}{\sqrt3},\qquad r=\frac{a}{2\sqrt3}$$ $$R=2r,\qquad \text{height}=\frac{\sqrt3}{2}a=3r$$

and the circumcentre, incentre, centroid and orthocentre are the same point. In terms of $R$:

$$\Delta=\frac{3\sqrt3}{4}R^2,\qquad a=R\sqrt3$$

That last pair is what a question means when it gives you the area of an equilateral triangle inscribed in a circle.

Example 23 (JEE Main 2026, 21 January Shift 1). A point $A$ lies between two parallel lines $L_1$ and $L_2$, at distances $6$ and $3$ from them. Find the area of the equilateral triangle $ABC$ with $B$ on $L_1$ and $C$ on $L_2$.

▸ SHOW SOLUTION

Equilateral triangle with one vertex between two parallel lines

Step 1. Set up one angle and use it twice

Let the side be $a$ and let $AC$ make an angle $\theta$ with the parallel lines. The perpendicular from $A$ to $L_2$ has length $3$, so

$$a\sin\theta=3$$

The side $AB$ makes an angle $60^\circ+\theta$ with the same direction, and $B$ lies on $L_1$, at perpendicular distance $6$ on the other side, so the total drop is $6+3=9$:

$$a\sin(60^\circ+\theta)=9$$

Step 2. Expand the second equation

$$a\left(\frac{\sqrt3}{2}\cos\theta+\frac12\sin\theta\right)=9$$$$\sqrt3\,a\cos\theta+a\sin\theta=18$$

Step 3. Substitute the first equation

$$\sqrt3\,a\cos\theta=15$$$$a\cos\theta=5\sqrt3$$

Step 4. Square and add

$$a^2=(a\sin\theta)^2+(a\cos\theta)^2$$$$a^2=9+75=84$$

Step 5. Area

$$\Delta=\frac{\sqrt3}{4}a^2=\frac{\sqrt3}{4}\cdot 84=21\sqrt3$$

Answer: (3) $21\sqrt3$.

Squaring and adding is what makes this quick. Solving for $\theta$ first produces an ugly inverse tangent and the same answer three lines later.

Example 24 (JEE Main 2025, 22 January Shift 2). Two parallel lines are $5$ apart and a point $P$ lies between them at distance $1$ from one. An equilateral triangle $PQR$ has $Q$ on one line and $R$ on the other. Find $(QR)^2$.

▸ SHOW SOLUTION

Step 1. Put in coordinates

Take the lines $y=0$ and $y=5$, and $P=(0,1)$. Let $Q=(q,0)$ and $R=(t,5)$.

Step 2. Write the three equal squared sides

$$PQ^2=q^2+1$$$$PR^2=t^2+16$$$$QR^2=(q-t)^2+25$$

Step 3. Solve $PQ^2=PR^2=QR^2$

From the first two, $q^2-t^2=15$. Substituting into $PQ^2=QR^2$ and simplifying gives

$$q=3\sqrt3,\qquad t=2\sqrt3$$

up to the reflection that swaps signs.

Step 4. Read off the answer

$$QR^2=\left(3\sqrt3-2\sqrt3\right)^2+25$$$$QR^2=3+25=28$$

Answer: $28$.

Check: $PQ^2=27+1=28$ and $PR^2=12+16=28$, so all three sides do agree.

Example 25 (JEE Main 2026, 6 April Shift 1). The centre of the circle $x^2+y^2+2gx+2fy+25=0$ lies in the first quadrant and on the line $2x-y=4$. An equilateral triangle inscribed in the circle has area $27\sqrt3$. Find the square of the length of the chord cut on $x=1$.

▸ SHOW SOLUTION

Equilateral triangle inscribed in the circle, and the chord on x = 1

Step 1. Turn the area into $R$

For an equilateral triangle inscribed in a circle of radius $R$, the side is $R\sqrt3$, so

$$\Delta=\frac{\sqrt3}{4}\left(R\sqrt3\right)^2=\frac{3\sqrt3}{4}R^2$$$$\frac{3\sqrt3}{4}R^2=27\sqrt3\implies R^2=36$$

Step 2. Use the two conditions on the centre

The centre is $(-g,-f)$, and it lies on $2x-y=4$:

$$-2g+f=4$$

The radius condition is $g^2+f^2-25=36$, so $g^2+f^2=61$. Substituting $f=4+2g$:

$$5g^2+16g-45=0$$$$g=\frac{-16\pm 34}{10}=\frac95 \ \text{or}\ -5$$

Step 3. Pick the root that puts the centre in the first quadrant

$g=-5$ gives the centre $(5,6)$, which qualifies. $g=\frac95$ gives a centre in the third quadrant, rejected.

Step 4. Cut the circle with $x=1$

$$(1-5)^2+(y-6)^2=36$$$$(y-6)^2=20$$

The chord has length $2\sqrt{20}$, so its square is

$$4\cdot 20=80$$

Answer: $80$.

Example 26 (JEE Main 2026, 5 April Shift 1). In an equilateral triangle $PQR$ the vertex $P$ is $(3,5)$ and the side $QR$ lies along $x+y=4$. If the orthocentre is $(\alpha,\beta)$, find $9(\alpha+\beta)$.

▸ SHOW SOLUTION

Step 1. The orthocentre is the centroid

In an equilateral triangle all four centres coincide, so it is enough to find the centroid.

Step 2. Find the foot of the perpendicular from $P$

$$M=P-\frac{x_P+y_P-4}{1^2+1^2}(1,1)$$$$M=(3,5)-\frac{4}{2}(1,1)=(1,3)$$

Step 3. Use $Q+R=2M$

$$G=\frac{P+Q+R}{3}=\frac{P+2M}{3}$$$$G=\left(\frac{3+2}{3},\ \frac{5+6}{3}\right)=\left(\frac53,\frac{11}{3}\right)$$

Step 4. Combine

$$9(\alpha+\beta)=9\cdot\frac{16}{3}=48$$

Answer: (4) $48$.

$Q$ and $R$ were never located, and never needed to be. Anything symmetric in $Q$ and $R$ depends only on their midpoint.

Example 27 (JEE Main 2026, 28 January Shift 1). $ABC$ is equilateral with orthocentre at the origin and side $BC$ on the line $x+2\sqrt2\,y=4$. If $A=(\alpha,\beta)$, find $\left\lfloor\left|\alpha+\sqrt2\,\beta\right|\right\rfloor$.

▸ SHOW SOLUTION

Equilateral triangle with its orthocentre at the origin

Step 1. Locate the foot of the altitude

The orthocentre is the centroid $O(0,0)$. The distance from $O$ to $BC$ is

$$OD=\frac{|0+0-4|}{\sqrt{1+8}}=\frac43$$

Step 2. Use the $2:1$ split

$$AO=2\,OD=\frac83$$$$AD=AO+OD=4$$

Step 3. Step from $O$ away from the line

The unit normal towards the line is $\dfrac{1}{3}\left(1,2\sqrt2\right)$, so $A$ is at $-\dfrac83$ times it:

$$A=-\frac{8}{9}\left(1,2\sqrt2\right)=\left(-\frac89,\ -\frac{16\sqrt2}{9}\right)$$

Step 4. Evaluate the required expression

$$\sqrt2\,\beta=\sqrt2\cdot\left(-\frac{16\sqrt2}{9}\right)=-\frac{32}{9}$$$$\alpha+\sqrt2\,\beta=-\frac89-\frac{32}{9}=-\frac{40}{9}$$$$\left\lfloor\frac{40}{9}\right\rfloor=\left\lfloor 4.44\ldots\right\rfloor=4$$

Answer: (4) $4$.

The sign matters. $A$ must lie on the opposite side of $BC$ from the direction the normal points towards the line, and taking the other sign gives $\alpha=\frac{16}{9}$, which puts $A$ on the same side as $BC$ and produces a wrong option that is on the paper.

Angle bisectors and the isosceles trap

The internal bisector from $B$ meets $AC$ at $D$ with

$$\frac{AD}{DC}=\frac{BA}{BC}=\frac{c}{a}$$

and the external bisector meets $AC$ produced at $D'$ with the same ratio taken externally. Its length is

$$BD=\frac{2ac\cos\frac{B}{2}}{a+c}$$

Proof of the ratio. Areas of $\triangle ABD$ and $\triangle DBC$ share the same height from $B$, so their ratio is $AD:DC$. Computing them instead as $\frac12\cdot BA\cdot BD\sin\frac{B}{2}$ and $\frac12\cdot BD\cdot BC\sin\frac{B}{2}$ gives the ratio $BA:BC$.

Example 28 (JEE Main 2026, 24 January Shift 1). $A(1,0)$, $B(2,-1)$ and $C\left(\dfrac73,\dfrac43\right)$ are three points. If the bisector of $\angle ABC$ is $\alpha x+\beta y=5$, find $\alpha^2+\beta^2$.

▸ SHOW SOLUTION

Step 1. Find the two sides at $B$

$$BA=\sqrt{1+1}=\sqrt2$$$$BC=\sqrt{\left(\tfrac13\right)^2+\left(\tfrac73\right)^2}=\sqrt{\frac{50}{9}}=\frac{5\sqrt2}{3}$$

Step 2. Use the bisector ratio

$$\frac{AD}{DC}=\frac{BA}{BC}=\frac{\sqrt2}{5\sqrt2/3}=\frac35$$

Step 3. Section formula on $AC$

$D$ divides $AC$ in the ratio $3:5$, so

$$D_x=\frac{3\cdot\frac73+5\cdot 1}{8}=\frac{12}{8}=\frac32$$$$D_y=\frac{3\cdot\frac43+5\cdot 0}{8}=\frac48=\frac12$$

Step 4. Write the line $BD$

$$\text{slope}=\frac{\frac12+1}{\frac32-2}=\frac{3/2}{-1/2}=-3$$$$y+1=-3(x-2)\implies 3x+y=5$$

so $\alpha=3$, $\beta=1$.

$$\alpha^2+\beta^2=9+1=10$$

Answer: (4) $10$.

The form $\alpha x+\beta y=5$ fixes the scaling: once you have $3x+y=5$ there is nothing left to normalise. Had the line come out as $6x+2y=10$ you would have to divide by $2$ before reading off $\alpha$ and $\beta$.

The internal bisector from B meeting AC at D

Example 29 (JEE Main 2025, 28 January Shift 2). The two equal sides of an isosceles triangle lie along $-x+2y=4$ and $x+y=4$. If $m$ is the slope of the third side, find the sum of all possible distinct values of $m$.

▸ SHOW SOLUTION

Step 1. Say what "equal sides" gives you

The base makes equal angles with the two equal sides, so with $m_1=\dfrac12$ and $m_2=-1$,

$$\left|\frac{m-m_1}{1+mm_1}\right|=\left|\frac{m-m_2}{1+mm_2}\right|$$

Step 2. Square to remove the moduli

$$\left(\frac{2m-1}{2+m}\right)^2=\left(\frac{m+1}{1-m}\right)^2$$

Step 3. Take both signs

The sign giving $(2m-1)(1-m)=(m+1)(2+m)$ leads to $3m^2=-3$, which has no real root. The other sign gives

$$(2m-1)(1-m)=-(m+1)(2+m)$$$$m^2-6m-1=0$$

Step 4. Sum the roots

$$m=3\pm\sqrt{10},\qquad \text{sum}=6$$

Answer: (3) $6$.

Both roots are genuine: they are the base of the isosceles triangle and the base of its reflection, and the question asks for the sum, so Vieta's relation gives the answer without ever computing $\sqrt{10}$.

Cyclic quadrilaterals

A quadrilateral inscribed in a circle has opposite angles supplementary, and that one fact does everything.

Cyclic quadrilateral with its two diagonals and Ptolemy's relation

With sides $AB=a$, $BC=b$, $CD=c$, $DA=d$ and $s=\dfrac{a+b+c+d}{2}$:

$$B+D=180^\circ$$ $$\cos B=\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}$$ $$\Delta=\frac12(ab+cd)\sin B$$ $$\Delta=\sqrt{(s-a)(s-b)(s-c)(s-d)}$$ $$AC\cdot BD=ac+bd$$

Proof of the cosine formula. Apply the cosine rule to $AC$ in $\triangle ABC$ and in $\triangle ACD$:

$$AC^2=a^2+b^2-2ab\cos B$$ $$AC^2=c^2+d^2-2cd\cos D$$

Since $D=180^\circ-B$, $\cos D=-\cos B$. Equating the two and solving for $\cos B$ gives the formula. The area formula follows by adding the two triangle areas and using $\sin D=\sin B$. The last line is Ptolemy's theorem; the fourth is Brahmagupta's formula, which is Heron's formula with a fourth factor.

For a general quadrilateral, not necessarily cyclic, the area is

$$\Delta=\sqrt{P-abcd\cos^2\alpha}$$

where $P=(s-a)(s-b)(s-c)(s-d)$ and $\alpha=\dfrac{A+C}{2}$. Setting $A+C=180^\circ$ recovers Brahmagupta, which shows that the cyclic quadrilateral is the one of largest area for given side lengths.

Example 30 (IITianForum). A cyclic quadrilateral has sides $3$, $4$, $5$, $6$ in order. Find its area, $\cos B$, and the product of its diagonals.

▸ SHOW SOLUTION

Cyclic quadrilateral with sides 3, 4, 5, 6

Step 1. Semi-perimeter and Brahmagupta

$$s=\frac{3+4+5+6}{2}=9$$$$\Delta=\sqrt{6\cdot 5\cdot 4\cdot 3}=\sqrt{360}=6\sqrt{10}$$

Step 2. The angle between the sides $3$ and $4$

$$\cos B=\frac{9+16-25-36}{2(12+30)}=\frac{-36}{84}=-\frac37$$

so $B$ is obtuse, as it must be when the two opposite sides are the longer pair.

Step 3. Cross-check the area

$$\sin B=\sqrt{1-\frac{9}{49}}=\frac{2\sqrt{10}}{7}$$$$\Delta=\frac12(12+30)\cdot\frac{2\sqrt{10}}{7}$$$$\Delta=21\cdot\frac{2\sqrt{10}}{7}=6\sqrt{10}$$

which agrees with Step 1.

Step 4. Ptolemy

$$AC\cdot BD=3\cdot 5+4\cdot 6=39$$

Answer: $\Delta=6\sqrt{10}$, $\cos B=-\dfrac37$, $AC\cdot BD=39$.

Regular polygons

A regular polygon of $n$ sides is $n$ copies of one isosceles triangle joined at the centre, and every formula follows from solving that one triangle.

Regular polygon split into n congruent triangles about the centre

$$\text{sum of interior angles}=(n-2)\pi$$ $$\text{each interior angle}=\frac{(n-2)\pi}{n}$$ $$R=\frac{a}{2}\operatorname{cosec}\frac{\pi}{n},\qquad r=\frac{a}{2}\cot\frac{\pi}{n}$$ $$\text{area}=\frac14 na^2\cot\frac{\pi}{n}$$ $$\text{area}=nr^2\tan\frac{\pi}{n}=\frac{n}{2}R^2\sin\frac{2\pi}{n}$$

Proof. The central angle subtended by one side is $\dfrac{2\pi}{n}$. Half of that isosceles triangle is right angled with hypotenuse $R$, opposite side $\dfrac{a}{2}$ and adjacent side $r$, giving the two radius formulas at once. The area is $n$ times $\dfrac12 R^2\sin\dfrac{2\pi}{n}$.

Example 31 (IITianForum). A regular polygon inscribed in a circle of radius $R$ has area $3R^2$. Find the number of sides.

▸ SHOW SOLUTION

A regular 12-gon with one of the twelve congruent triangles shaded

Step 1. Use the area formula in terms of $R$

$$\frac{n}{2}R^2\sin\frac{2\pi}{n}=3R^2$$$$\frac{n}{2}\sin\frac{2\pi}{n}=3$$

Step 2. Try the value that makes the angle standard

$n=12$ gives $\dfrac{2\pi}{n}=30^\circ$, so

$$\frac{12}{2}\sin 30^\circ=6\cdot\frac12=3$$

which works.

Step 3. Confirm there is no other

$f(n)=\dfrac{n}{2}\sin\dfrac{2\pi}{n}$ is strictly increasing in $n$, since it is the area of the inscribed $n$-gon in a fixed circle and adding a side strictly increases that area. So the solution is unique.

Answer: $n=12$.

As a check on the direction of that monotonicity, $f(n)\to\pi$ as $n\to\infty$, which is the area of the unit circle, and $3<\pi$, so a finite answer had to exist.

Which rule to reach for

Decision chart: what you are given decides which rule to use

Read it in one line: what you are given decides the rule, and what you are asked decides which area or radius formula finishes the job. The half angle formulae sit between the two, because they are the only tool that converts an angle condition into a pure statement about $s$, $a$, $b$ and $c$.

Quick recall

Practise these until the decision chart is automatic, then take a full chapter test in the doMath practice section to see how fast you can place a question in the right box.