Difference between revisions of "2023 IMO Problems/Problem 2"

Latest revision as of 09:58, 23 August 2024

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 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.

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. $\blacksquare$ ~approved by Kislay kai

@@ Line 1: / Line 1: @@
 ==Problem==
-[[File:2023 IMO 2o.png|250px|right]]
 Let <math>ABC</math> be an acute-angled triangle with <math>AB < AC</math>. Let <math>\Omega</math> be the circumcircle of <math>ABC</math>. Let <math>S</math> be the midpoint of the arc <math>CB</math> of <math>\Omega</math> containing <math>A</math>. The perpendicular from <math>A</math> to <math>BC</math> meets <math>BS</math> at <math>D</math> and meets <math>\Omega</math> again at <math>E \neq A</math>. The line through <math>D</math> parallel to <math>BC</math> meets line <math>BE</math> at <math>L</math>. Denote the circumcircle of triangle <math>BDL</math> by <math>\omega</math>. Let <math>\omega</math> meet <math>\Omega</math> again at <math>P \neq B</math>. Prove that the line tangent to <math>\omega</math> at <math>P</math> meets line <math>BS</math> on the internal angle bisector of <math>\angle BAC</math>.
-==Solutions==
+==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==
+[[File:2023 IMO 2o.png|250px|right]]
 [[File:2023 IMO 2o0.png|400px|right]]
 Denote the point diametrically opposite to a point <math>S</math> through <math>S' \implies AS'</math> is the internal angle bisector of <math>\angle BAC</math>.
@@ Line 23: / Line 28: @@
 Denote <math>T = S'B \cap SA \implies SB \perp TS', S'A \perp TS \implies H</math> is the orthocenter of <math>\triangle TSS'.</math>
-Denote <math>G = PS' \cap AE \implies \angle BPG = \angle BPS' = \angle BSS' = \angle BDG \implies B, L, P, D,</math> and <math>G</math> are concyclic.
+Denote <math>G = PS' \cap AE \implies \angle BPG = \angle BPS' = \angle BSS' = \angle BDG</math> <math>\implies B, L, P, D,</math> and <math>G</math> are concyclic.
 [[File:2023 IMO 2 lemma.png|300px|right]]
 <math>\angle EBS' = \varphi, \angle LBG = \angle LDG = 90^\circ = \angle DBS' \implies \angle DBG = \varphi = \angle SBF \implies</math>
@@ Line 63: / Line 68: @@
 '''vladimir.shelomovskii@gmail.com, vvsss'''
+==Solution 2==
+Let <math>O</math> be the circumcenter of <math>\triangle ABC</math>. We proceed with showing that <math>PH=AH</math>. Suppose that <math>PD</math> intersects <math>SS'</math> and <math>\Omega</math> at <math>Q</math> and <math>F \ne A</math> respectively. Note that
+<cmath>\angle CBE=\angle DLE=\angle DPB=\angle FCB \implies EF \| BC.</cmath>
+Since <math>AE \perp BC</math>, we have <math>AE \perp EF</math> and hence <math>AF</math> is a diameter of <math>\Omega</math>. By similar triangles <math>OQ=\frac{1}{2}AD</math> and therefore
+<cmath>\frac{SO}{OQ}=\frac{SS'}{AD}=\frac{SH}{HD} \implies OH \| DQ.</cmath>
+Since <math>AF</math> is a diameter of <math>\Omega</math>, <math>FP \perp AP \implies OH \perp AP</math> and thus <math>H</math> lies on the perpendicular bisector of <math>AP</math>. This proves the claim.
+==Solution 3==
+Identify <math>\Omega</math> with the unit circle, and let the internal bisector of <math>\angle BAC</math> meet <math>\overleftrightarrow{BS}</math> at <math>Q</math> and <math>\omega</math> again at <math>T</math>. 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 <math>P</math>. We have <math>|p|=1</math>, and
+<cmath>\frac{(b-p)(d-\ell)}{(b-\ell)(d-p)} \in \mathbb{R}</cmath>
+Now <math>\frac{d-\ell}{b-c}, \frac{b-\ell}{b-e} \in \mathbb{R}</math>. Thus we have
+<cmath>\frac{(b-p)(b-c)}{(b-e)(d-p)} \in \mathbb{R}</cmath>
+and so
+<cmath>\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)}</cmath>
+<cmath>a(ps^2+aps+abp-a^2p-as^2-abs) = s(a^2b+abs+as^2-bs^2-abp-aps)</cmath>
+<cmath>2aps^2+a^2ps+a^2bp-a^3p-a^2s^2-2a^2bs-abs^2-as^3+bs^3+abps = 0</cmath>
+<cmath>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)}</cmath>
+It remains to show that <math>Q</math> lies on the tangent to <math>\omega</math> at <math>P</math>. Now let <math>O</math> be the center of <math>\omega</math>. Define the vectors
+<cmath>b' = b-p = \frac{(b-a)(ab+s^2)}{a(2s+b-a)}</cmath>
+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 <math>\overline{b'} = -\frac{b'}{bp}</math> and <math>\overline{d'} = \frac{d'}{ap}</math>. Thus if we define <math>o' = o-p</math>, we have
+<cmath>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)}</cmath>
+Meanwhile, we compute
+<cmath>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)}</cmath>
+So
+<cmath>\frac{o-p}{q-p} = \frac{(ab+s^2)}{(a-s)(b+s)}</cmath>
+which is pure imaginary. <math>\blacksquare</math>
+~approved by Kislay kai
+==See Also==
+{{IMO box|year=2023|num-b=1|num-a=3}}

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