Difference between revisions of "2023 USAJMO Problems/Problem 2"
(→Solution 2) |
|||
Line 23: | Line 23: | ||
- Leo.Euler | - Leo.Euler | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | We are going to use barycentric coordinates on <math>\triangle ABC</math>. Let <math>A=(1,0,0)</math>, <math>B=(0,1,0)</math>, <math>C=(0,0,1)</math>, and <math>a=BC</math>, <math>b=CA</math>, <math>c=AB</math>. We have <math>M=\left(0,\frac{1}{2},\frac{1}{2}\right)</math> and <math>P=(x:1:1)</math> so <math>\overrightarrow{CP}=\left(\frac{x}{x+2},\frac{1}{x+2},\frac{1}{x+2}-1\right)</math> and <math>\overrightarrow{AM}=\left(-1,\frac{1}{2},\frac{1}{2}\right)</math>. Since <math>\overleftrightarrow{CP}\perp\overleftrightarrow{AM}</math>, it follows that | ||
+ | \begin{align*} | ||
+ | a^2\left(\frac{1}{2}\cdot\frac{1}{x+2}+\frac{1}{2}\left(\frac{1}{x+2}-1\right)\right)+b^2\left(\frac{1}{2}\cdot\frac{x}{x+2}-\left(\frac{1}{x+2}-1\right)\right)\\ | ||
+ | +c^2\left(\frac{1}{2}\cdot\frac{x}{x+2}-\frac{1}{x+2}\right)=0. | ||
+ | \end{align*}Solving this gives | ||
+ | \[ | ||
+ | x=\frac{2b^2-2c^2}{a^2-3b^2-c^2} | ||
+ | \]so | ||
+ | \[ | ||
+ | P=\left(\frac{b^2-c^2}{a^2-2b^2-2c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right). | ||
+ | \]The equation for <math>(ABP)</math> is | ||
+ | \[ | ||
+ | -a^2yz-b^2zx-c^2xy+ux+vy+wz=0. | ||
+ | \]Plugging in <math>A</math> and <math>B</math> gives <math>u=v=0</math>. Plugging in <math>P</math> gives | ||
+ | \begin{align*} | ||
+ | -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{align*}so | ||
+ | \[ | ||
+ | w=\frac{2b^4-2c^4+a^4-3a^2b^2-a^2c^2}{2a^2-4b^2-4c^2}=\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{2}. | ||
+ | \]Now let <math>Q=(0,t,1-t)</math> where | ||
+ | \begin{align*} | ||
+ | -a^2t(1-t)+w(1-t)&=0\\ | ||
+ | \implies t&=\frac{w}{a^2} | ||
+ | \end{align*}so <math>Q=\left(0,\frac{w}{a^2},1-\frac{w}{a^2}\right)</math>. It follows that <math>N=\left(\frac{1}{2},\frac{w}{2a^2},1-\frac{w}{2a^2}\right)</math>. It suffices to prove that <math>\overleftrightarrow{ON}\perp\overleftrightarrow{BC}</math>. Setting <math>\overrightarrow{O}=0</math>, we get <math>\overrightarrow{N}=\left(\frac{1}{2},\frac{w}{2a^2},1-\frac{w}{2a^2}\right)</math>. Furthermore we have <math>\overrightarrow{CB}=(0,1,-1)</math> so it suffices to prove that | ||
+ | \begin{align*} | ||
+ | a^2\left(-\frac{w}{2a^2}+\frac{1}{2}-\frac{u}{2a^2}\right)+b^2\left(-\frac{1}{2}\right)+c^2\left(\frac{1}{2}\right)=0\\ | ||
+ | \implies w=\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{2} | ||
+ | \end{align*} | ||
+ | which is valid. <math>\square</math> | ||
+ | |||
+ | ~KevinYang2.71 |
Revision as of 22:07, 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 \begin{align*} a^2\left(\frac{1}{2}\cdot\frac{1}{x+2}+\frac{1}{2}\left(\frac{1}{x+2}-1\right)\right)+b^2\left(\frac{1}{2}\cdot\frac{x}{x+2}-\left(\frac{1}{x+2}-1\right)\right)\\ +c^2\left(\frac{1}{2}\cdot\frac{x}{x+2}-\frac{1}{x+2}\right)=0. \end{align*}Solving this gives \[ x=\frac{2b^2-2c^2}{a^2-3b^2-c^2} \]so \[ P=\left(\frac{b^2-c^2}{a^2-2b^2-2c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right). \]The equation for is \[ -a^2yz-b^2zx-c^2xy+ux+vy+wz=0. \]Plugging in and gives . Plugging in gives \begin{align*} -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{align*}so \[ w=\frac{2b^4-2c^4+a^4-3a^2b^2-a^2c^2}{2a^2-4b^2-4c^2}=\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{2}. \]Now let where \begin{align*} -a^2t(1-t)+w(1-t)&=0\\ \implies t&=\frac{w}{a^2} \end{align*}so . It follows that . It suffices to prove that . Setting , we get . Furthermore we have so it suffices to prove that \begin{align*} a^2\left(-\frac{w}{2a^2}+\frac{1}{2}-\frac{u}{2a^2}\right)+b^2\left(-\frac{1}{2}\right)+c^2\left(\frac{1}{2}\right)=0\\ \implies w=\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{2} \end{align*} which is valid.
~KevinYang2.71