Difference between revisions of "2023 IMO Problems/Problem 2"
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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>. | ||
+ | |||
+ | ==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== | ==Solution== | ||
− | + | ||
+ | [[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>. | ||
+ | |||
+ | Denote the crosspoint of <math>BS</math> and <math>AS'</math> through <math>H, \angle ABS = \varphi.</math> | ||
+ | |||
+ | <cmath>AE \perp BC, SS' \perp BC \implies \overset{\Large\frown} {AS} = \overset{\Large\frown} {ES'} = 2\varphi \implies</cmath> | ||
+ | |||
+ | <cmath>\angle EAS' = \varphi = \angle ABS \implies \angle DAH = \angle ABH \implies</cmath> | ||
+ | <cmath>\triangle AHD \sim \triangle BAH \implies \frac {AH}{BH} = \frac {DH}{AH} \implies AH^2 = BH \cdot DH.</cmath> | ||
+ | To finishing the solution we need only to prove that <math>PH = AH.</math> | ||
+ | |||
+ | Denote <math>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} =</math> | ||
+ | <math>=\angle FCB + \varphi \implies \angle FBS = \angle ABS \implies H</math> is incenter of <math>\triangle ABF.</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. | ||
+ | [[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> | ||
+ | |||
+ | points <math>B, G,</math> and <math>F</math> are collinear <math>\implies GF</math> is symmetric to <math>AF</math> with respect <math>TF.</math> | ||
+ | |||
+ | We use the lemma and complete the proof. | ||
+ | |||
+ | <i><b>Lemma 1</b></i> | ||
+ | |||
+ | [[File:2023 IMO 2b Lemma.png|300px|right]] | ||
+ | Let acute triangle <math>\triangle ABC, AB > AC</math> be given. | ||
+ | |||
+ | Let <math>H</math> be the orthocenter of <math>\triangle ABC, BHD</math> be the height. | ||
+ | |||
+ | Let <math>\Omega</math> be the circle <math>BCD. BC</math> is the diameter of <math>\Omega.</math> | ||
+ | |||
+ | The point <math>E</math> is symmetric to <math>D</math> with respect to <math>AH.</math> | ||
+ | |||
+ | The line <math>BE</math> meets <math>\Omega</math> again at <math>F \neq B</math>. | ||
+ | |||
+ | Prove that <math>HF = HD.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>\omega</math> be the circle centered at <math>H</math> with radius <math>HD.</math> | ||
+ | |||
+ | The <math>\omega</math> meets <math>\Omega</math> again at <math>F' \neq D, HD = HF'.</math> | ||
+ | |||
+ | Let <math>\omega</math> meets <math>BF'</math> again at <math>E' \neq F'</math>. | ||
+ | |||
+ | We use Reim’s theorem for <math>\omega, \Omega</math> and lines <math>CDD</math> and <math>BE'F'</math> and get <math>E'D || BC</math> | ||
+ | |||
+ | (this idea was recommended by <i><b>Leonid Shatunov</b></i>). | ||
+ | |||
+ | <math>AH \perp BC \implies AH \perp E'D \implies</math> | ||
+ | |||
+ | The point <math>E'</math> is symmetric to <math>D</math> with respect to <math>AH \implies E' = E \implies F' = F \implies HF = HD.</math> | ||
+ | |||
+ | '''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 10:48, 4 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
\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
. We have
, and
Now
. Thus we have
and so
It remains to show that
lies on the tangent to
at
. Now let
be the center of
. Define the vectors
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
and
. Thus if we define
, we have
Meanwhile, we compute
So
which is pure imaginary.
~approved by Kislay kai
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 |