2023 USAJMO Problems/Problem 2
Contents
Problem
(Holden Mui) In an acute triangle , let be the midpoint of . Let be the foot of the perpendicular from to . Suppose that the circumcircle of triangle intersects line at two distinct points and . Let be the midpoint of . Prove that .
Video Solution
https://www.youtube.com/watch?v=f-d4mi-AyxQ
Solution 1
The condition is solved only if is isosceles, which in turn only happens if is perpendicular to .
Now, draw the altitude from to , and call that point . Because of the Midline Theorem, the only way that this condition is met is if , or if .
By similarity, . Using similarity ratios, we get that . Rearranging, we get that . This implies that is cyclic.
Now we start using Power of a Point. We get that , and from before. This leads us to get that .
Now we assign variables to the values of the segments. Let and . The equation from above gets us that . As from the problem statements, this gets us that and , and we are done.
-dragoon and rhydon516 (:
Solution 2
Let be the foot of the altitude from onto . We want to show that for obvious reasons.
Notice that is cyclic and that lies on the radical axis of and . By Power of a Point, . As , we have , as desired.
- Leo.Euler
Solution 3 (Less technical bary)
We are going to use barycentric coordinates on . Let , , , and , , . We have and so and . Since , it follows that Solving this gives so The equation for is Plugging in and gives . Plugging in gives so Now let where so . It follows that . It suffices to prove that . Setting , we get . Furthermore we have so it suffices to prove that which is valid.
~KevinYang2.71
Solution 4 (Less bashy bary)
We employ barycentric coordinates. Set as the reference triangle with , , and . We immediately have, Since it passes through , for some , the equation of circle is, Plugging in , Plugging in , In conclusion the circle has formula, is the second intersection of circle with . We let for some . Plugging this in, We claim that is the other solution. Factoring out the , this is clearly true.
We also check that, these are not the same value. The triangle is acute, so this is impossible.
Since we had a quadratic in with at most two solutions, the second intersection is indeed, Therefore,
~ Daniel Ge
See Also
2023 USAJMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
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