Difference between revisions of "2023 IMO Problems/Problem 2"
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==Problem== | ==Problem== | ||
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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>. | 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>. | ||
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==Solution== | ==Solution== | ||
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+ | [[File:2023 IMO 2o.png|250px|right]] | ||
[[File:2023 IMO 2o0.png|400px|right]] | [[File:2023 IMO 2o0.png|400px|right]] | ||
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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>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]] | [[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> | <math>\angle EBS' = \varphi, \angle LBG = \angle LDG = 90^\circ = \angle DBS' \implies \angle DBG = \varphi = \angle SBF \implies</math> | ||
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''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}} |
Latest revision as of 09:58, 23 August 2024
Problem
Let be an acute-angled triangle with . Let be the circumcircle of . Let be the midpoint of the arc of containing . The perpendicular from to meets at and meets again at . The line through parallel to meets line at . Denote the circumcircle of triangle by . Let meet again at . Prove that the line tangent to at meets line on the internal angle bisector of .
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 through is the internal angle bisector of .
Denote the crosspoint of and through
To finishing the solution we need only to prove that
Denote is incenter of
Denote is the orthocenter of
Denote and are concyclic.
points and are collinear is symmetric to with respect
We use the lemma and complete the proof.
Lemma 1
Let acute triangle be given.
Let be the orthocenter of be the height.
Let be the circle is the diameter of
The point is symmetric to with respect to
The line meets again at .
Prove that
Proof
Let be the circle centered at with radius
The meets again at
Let meets again at .
We use Reim’s theorem for and lines and and get
(this idea was recommended by Leonid Shatunov).
The point is symmetric to with respect to
vladimir.shelomovskii@gmail.com, vvsss
Solution 2
Let be the circumcenter of . We proceed with showing that . Suppose that intersects and at and respectively. Note that
Since , we have and hence is a diameter of . By similar triangles and therefore
Since is a diameter of , and thus lies on the perpendicular bisector of . This proves the claim.
Solution 3
Identify with the unit circle, and let the internal bisector of meet at and again at . We set up so that
See Also
2023 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |