Difference between revisions of "2023 USAJMO Problems/Problem 2"
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\]</cmath> | \]</cmath> | ||
Plugging in <math>A</math> and <math>B</math> gives <math>u=v=0</math>. Plugging in <math>P</math> gives | Plugging in <math>A</math> and <math>B</math> gives <math>u=v=0</math>. Plugging in <math>P</math> gives | ||
− | <cmath>\begin{ | + | <cmath>\begin{multline*} |
-a^2\left(\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right)^2-b^2\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\cdot\frac{b^2-c^2}{a^2-2b^2-2c^2}\\ | -a^2\left(\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right)^2-b^2\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\cdot\frac{b^2-c^2}{a^2-2b^2-2c^2}\\ | ||
-c^2\cdot\frac{b^2-c^2}{a^2-2b^2-2c^2}\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}+w\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}=0 | -c^2\cdot\frac{b^2-c^2}{a^2-2b^2-2c^2}\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}+w\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}=0 | ||
− | \end{ | + | \end{multline*}</cmath> |
so | so | ||
<cmath>\[ | <cmath>\[ |
Revision as of 22:14, 26 April 2023
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 .
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
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
\[\begin{multline*} -a^2\left(\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right)^2-b^2\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\cdot\frac{b^2-c^2}{a^2-2b^2-2c^2}\\ -c^2\cdot\frac{b^2-c^2}{a^2-2b^2-2c^2}\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}+w\cdot\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}=0 \end{multline*}\] (Error compiling LaTeX. Unknown error_msg)
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