2023 IMO Problems/Problem 2
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 |