Difference between revisions of "2023 USAJMO Problems/Problem 2"
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- Leo.Euler | - Leo.Euler | ||
− | ==Solution 3== | + | ==Solution 3 (Less technical bary) == |
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 | 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 | ||
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~KevinYang2.71 | ~KevinYang2.71 | ||
+ | |||
+ | == Solution 4 (Less bashy bary) == | ||
+ | We employ barycentric coordinates. Set <math>AMC</math> as the reference triangle with <math>A = (1, 0, 0)</math>, <math>M = (0, 1, 0)</math>, and <math>C = (0, 0, 1)</math>. We immediately have, | ||
+ | <cmath>P = (S_B : S_A : 0); B = (0, 2, -1)</cmath> | ||
+ | Since it passes through <math>A</math>, for some <math>v, w</math>, the equation of circle <math>(ABP)</math> is, | ||
+ | <cmath>(ABP): -a^2 yz - b^2 zx - c^2 xy + (vy + wz)(x + y + z) = 0</cmath> | ||
+ | Plugging in <math>P</math>, | ||
+ | <cmath>- c^2 S_{AB} + (v S_A)(c^2) = 0</cmath> | ||
+ | <cmath>\iff v = S_B</cmath> | ||
+ | Plugging in <math>B</math>, | ||
+ | <cmath>2a^2 + (2v - w) = 0</cmath> | ||
+ | <cmath>\iff 2a^2 + 2 S_B = 3 a^2 - b^2 + c^2 = w</cmath> | ||
+ | In conclusion the circle has formula, | ||
+ | <cmath>(ABP): -a^2 yz - b^2 zx - c^2 xy + ((S_B)y + (3 a^2 - b^2 + c^2) z)(x + y + z) = 0</cmath> | ||
+ | <math>Q</math> is the second intersection of circle <math>(ABP)</math> with <math>\overline{CM}</math>. We let <math>Q = (0, 1 - t, t)</math> for some <math>t \neq -1</math>. Plugging this in, | ||
+ | <cmath>-a^2 (1-t)t + ((S_B)(1-t) + (3 a^2 - b^2 + c^2) t) = 0</cmath> | ||
+ | We claim that <math>t = -\frac{S_B}{a^2}</math> is the other solution. | ||
+ | <cmath>\left(1+ \frac{S_B}{a^2} \right) S_B + \left((S_B)\left(1+ \frac{S_B}{a^2}\right) - (3 a^2 - b^2 + c^2) \frac{S_B}{a^2}\right) = 0</cmath> | ||
+ | <cmath>\iff \left(\frac{3a^2 - b^2 + c^2}{2a^2} \right) + \left(\left(\frac{3a^2 - b^2 + c^2}{2a^2}\right) - (3 a^2 - b^2 + c^2) \frac{1}{a^2}\right) = 0</cmath> | ||
+ | Factoring out the <math>\frac{3a^2 - b^2 + c^2}{2a^2}</math>, this is clearly true. | ||
+ | |||
+ | We also check that, these are not the same value. | ||
+ | <cmath>-\frac{S_B}{a^2} = -1</cmath> | ||
+ | <cmath>\iff a^2 - b^2 + c^2 = 2a^2</cmath> | ||
+ | <cmath>\iff c^2 = a^2 + b^2</cmath> | ||
+ | The triangle is acute, so this is impossible. | ||
+ | |||
+ | Since we had a quadratic in <math>t</math> with at most two solutions, the second intersection <math>Q</math> is indeed, | ||
+ | <cmath>Q = \left( 0, 1 + \frac{S_B}{a^2}, -\frac{S_B}{a^2} \right)</cmath> | ||
+ | Therefore, | ||
+ | <cmath>N = \frac{A + Q}{2} = \left( \frac{1}{2}, \frac{1}{2} - \frac{S_B}{2a^2}, \frac{S_B}{2a^2} \right)</cmath> | ||
+ | |||
==See Also== | ==See Also== | ||
{{USAJMO newbox|year=2023|num-b=1|num-a=3}} | {{USAJMO newbox|year=2023|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:22, 6 February 2024
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 (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,
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.