# 2023 IMO Problems/Problem 2

## Problem

Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.

## Video Solution

https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems]

https://youtu.be/I4FoXeVnSpM?si=BMxwsH4thsZ-7uLF [Video contains IMO 2023 P2 motivation + discussion] (~little-fermat)

## Solution

Denote the point diametrically opposite to a point $S$ through $S' \implies AS'$ is the internal angle bisector of $\angle BAC$.

Denote the crosspoint of $BS$ and $AS'$ through $H, \angle ABS = \varphi.$

$$AE \perp BC, SS' \perp BC \implies \overset{\Large\frown} {AS} = \overset{\Large\frown} {ES'} = 2\varphi \implies$$

$$\angle EAS' = \varphi = \angle ABS \implies \angle DAH = \angle ABH \implies$$ $$\triangle AHD \sim \triangle BAH \implies \frac {AH}{BH} = \frac {DH}{AH} \implies AH^2 = BH \cdot DH.$$ To finishing the solution we need only to prove that $PH = AH.$

Denote $F = SS' \cap AC \implies \angle CBS = \frac {\overset{\Large\frown} {CS}}{2} = \frac {\overset{\Large\frown} {BS}}{2} = \frac {\overset{\Large\frown} {AB}}{2} + \frac {\overset{\Large\frown} {AS}}{2} =$ $=\angle FCB + \varphi \implies \angle FBS = \angle ABS \implies H$ is incenter of $\triangle ABF.$

Denote $T = S'B \cap SA \implies SB \perp TS', S'A \perp TS \implies H$ is the orthocenter of $\triangle TSS'.$

Denote $G = PS' \cap AE \implies \angle BPG = \angle BPS' = \angle BSS' = \angle BDG \implies B, L, P, D,$ and $G$ are concyclic.

$\angle EBS' = \varphi, \angle LBG = \angle LDG = 90^\circ = \angle DBS' \implies \angle DBG = \varphi = \angle SBF \implies$

points $B, G,$ and $F$ are collinear $\implies GF$ is symmetric to $AF$ with respect $TF.$

We use the lemma and complete the proof.

Lemma 1

Let acute triangle $\triangle ABC, AB > AC$ be given.

Let $H$ be the orthocenter of $\triangle ABC, BHD$ be the height.

Let $\Omega$ be the circle $BCD. BC$ is the diameter of $\Omega.$

The point $E$ is symmetric to $D$ with respect to $AH.$

The line $BE$ meets $\Omega$ again at $F \neq B$.

Prove that $HF = HD.$

Proof

Let $\omega$ be the circle centered at $H$ with radius $HD.$

The $\omega$ meets $\Omega$ again at $F' \neq D, HD = HF'.$

Let $\omega$ meets $BF'$ again at $E' \neq F'$.

We use Reim’s theorem for $\omega, \Omega$ and lines $CDD$ and $BE'F'$ and get $E'D || BC$

(this idea was recommended by Leonid Shatunov).

$AH \perp BC \implies AH \perp E'D \implies$

The point $E'$ is symmetric to $D$ with respect to $AH \implies E' = E \implies F' = F \implies HF = HD.$

vladimir.shelomovskii@gmail.com, vvsss

## Solution 3

Identify $\Omega$ with the unit circle, and let the internal bisector of $\angle BAC$ meet $\overleftrightarrow{BS}$ at $Q$ and $\omega$ again at $T$. We set up so that \begin{align*} |a|=|b|=|s|&=1 \\ c &= \frac{s^2}b \\ t &= -s \\ e = -\frac{bc}a &= -\frac{s^2}a \\ d = \frac{ae(b+s) - bs(a+e)}{ae-bs} &= \frac{a^2b+abs+as^2-bs^2}{a(b+s)} \\ q = \frac{at(b+s) - bs(a+t)}{at-bs} &= \frac{2ab+as-bs}{a+b} \end{align*} Now we find the coordinate of $P$. We have $|p|=1$, and $$\frac{(b-p)(d-\ell)}{(b-\ell)(d-p)} \in \mathbb{R}$$ Now $\frac{d-\ell}{b-c}, \frac{b-\ell}{b-e} \in \mathbb{R}$. Thus we have $$\frac{(b-p)(b-c)}{(b-e)(d-p)} \in \mathbb{R}$$ and so $$\frac{a^2(b-p)(b+s)^2(b-s)}{b(ab+s^2)(a^2b+abs+as^2-bs^2-abp-aps)} = \frac{as(b-p)(b+s)^2(b-s)}{b(ab+s^2)(ps^2+aps+abp-a^2p-as^2-abs)}$$ $$a(ps^2+aps+abp-a^2p-as^2-abs) = s(a^2b+abs+as^2-bs^2-abp-aps)$$ $$2aps^2+a^2ps+a^2bp-a^3p-a^2s^2-2a^2bs-abs^2-as^3+bs^3+abps = 0$$ $$p = \frac{2a^2bs+abs^2+as^3+a^2s^2-bs^3}{2as^2+a^2s+a^2b+abs-a^3} = \frac{s(2ab+as-bs)}{a(2s+b-a)}$$ It remains to show that $Q$ lies on the tangent to $\omega$ at $P$. Now let $O$ be the center of $\omega$. Define the vectors $$b' = b-p = \frac{(b-a)(ab+s^2)}{a(2s+b-a)}$$ and \begin{align*} d' = d-p &= \frac{(2s+b-a)(a^2b+abs+as^2-bs^2) - s(b+s)(2ab+as-bs)}{a(b+s)(2s+b-a)} \\ &= \frac{a^2bs+as^3-bs^3+a^2b^2-a^3b-a^2s^2+abs^2-ab^2s}{a(b+s)(2s+b-a)} \\ &= \frac{(a-s)(b-a)(ab+s^2)}{a(b+s)(2s+b-a)} \end{align*} We observe that $\overline{b'} = -\frac{b'}{bp}$ and $\overline{d'} = \frac{d'}{ap}$. Thus if we define $o' = o-p$, we have $$o' = \frac{b'd'(\overline{b'}-\overline{d'})}{\overline{b'}d'-b'\overline{d'}} = \frac{ab'+bd'}{a+b} = \frac{\frac{(b-a)(ab+s^2)}{2s+b-a} + \frac{b(a-s)(b-a)(ab+s^2)}{a(b+s)(2s+b-a)}}{a+b} = \frac{(b-a)(ab+s^2)(2ab+as-bs)}{a(a+b)(b+s)(2s+b-a)}$$ Meanwhile, we compute $$q-p = \frac{2ab+as-bs}{a+b} - \frac{s(2ab+as-bs)}{a(2s+b-a)} = \frac{(a-s)(b-a)(2ab+as-bs)}{a(a+b)(2s+b-a)}$$ So $$\frac{o-p}{q-p} = \frac{(ab+s^2)}{(a-s)(b+s)}$$ which is pure imaginary. $\blacksquare$

## Solution 2

Let $O$ be the circumcenter of $\triangle ABC$. We proceed with showing that $PH=AH$. Suppose that $PD$ intersects $SS'$ and $\Omega$ at $Q$ and $F \ne A$ respectively. Note that

$$\angle CBE=\angle DLE=\angle DPB=\angle FCB \implies EF \| BC.$$

Since $AE \perp BC$, we have $AE \perp EF$ and hence $AF$ is a diameter of $\Omega$. By similar triangles $OQ=\frac{1}{2}AD$ and therefore

$$\frac{SO}{OQ}=\frac{SS'}{AD}=\frac{SH}{HD} \implies OH \| DQ.$$

Since $AF$ is a diameter of $\Omega$, $FP \perp AP \implies OH \perp AP$ and thus $H$ lies on the perpendicular bisector of $AP$. This proves the claim.

## See Also

 2023 IMO (Problems) • Resources Preceded byProblem 1 1 • 2 • 3 • 4 • 5 • 6 Followed byProblem 3 All IMO Problems and Solutions