Difference between revisions of "2023 USAMO Problems/Problem 1"
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Now, <math>NB = NC</math> iff <math>N</math> lies on the perpendicular bisector of <math>\overline{BC}</math>. As <math>N</math> lies on the perpendicular bisector of <math>\overline{XQ}</math>, which is also the perpendicular bisector of <math>\overline{BC}</math> (as <math>M</math> is also the midpoint of <math>XQ</math>), we are done. | Now, <math>NB = NC</math> iff <math>N</math> lies on the perpendicular bisector of <math>\overline{BC}</math>. As <math>N</math> lies on the perpendicular bisector of <math>\overline{XQ}</math>, which is also the perpendicular bisector of <math>\overline{BC}</math> (as <math>M</math> is also the midpoint of <math>XQ</math>), we are done. | ||
~ Martin2001 | ~ Martin2001 | ||
+ | |||
+ | ==Solution 2 == | ||
+ | |||
+ | 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 | ||
+ | <cmath>\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*}</cmath> | ||
+ | Solving this gives | ||
+ | <cmath>\[ | ||
+ | x=\frac{2b^2-2c^2}{a^2-3b^2-c^2} | ||
+ | \]</cmath> | ||
+ | so | ||
+ | <cmath>\[ | ||
+ | P=\left(\frac{b^2-c^2}{a^2-2b^2-2c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2}\right). | ||
+ | \]</cmath> | ||
+ | The equation for <math>(ABP)</math> is | ||
+ | <cmath>\[ | ||
+ | -a^2yz-b^2zx-c^2xy+ux+vy+wz=0. | ||
+ | \]</cmath> | ||
+ | Plugging in <math>A</math> and <math>B</math> gives <math>u=v=0</math>. Plugging in <math>P</math> gives | ||
+ | <cmath>\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*}</cmath> | ||
+ | so | ||
+ | <cmath>\[ | ||
+ | w=\frac{2b^4-2c^4+a^4-3a^2b^2-a^2c^2}{2a^2-4b^2-4c^2}=\frac{a^2}{2}-\frac{b^2}{2}+\frac{c^2}{2}. | ||
+ | \]</cmath> | ||
+ | Now let <math>Q=(0,t,1-t)</math> where | ||
+ | <cmath>\begin{align*} | ||
+ | -a^2t(1-t)+w(1-t)&=0\ | ||
+ | \implies t&=\frac{w}{a^2} | ||
+ | \end{align*}</cmath> | ||
+ | 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 | ||
+ | <cmath>\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*}</cmath> | ||
+ | which is valid. <math>\square</math> | ||
+ | |||
+ | ~KevinYang2.71 | ||
==See also== | ==See also== | ||
{{USAMO box|year=2023|before=First Problem|num-a=2|n=I}} | {{USAMO box|year=2023|before=First Problem|num-a=2|n=I}} |
Revision as of 18:17, 28 April 2023
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
Let be the foot from to . By definition, . Thus, , and .
From this, we have , as . Thus, is also the midpoint of .
Now, iff lies on the perpendicular bisector of . As lies on the perpendicular bisector of , which is also the perpendicular bisector of (as is also the midpoint of ), we are done. ~ Martin2001
Solution 2
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
See also
2023 USAMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |