Difference between revisions of "2023 USAJMO Problems/Problem 2"
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Now we start using Power of a Point. We get that <math>\overline{BX} \cdot \overline {XQ}= \overline{AM} \cdot \overline{MP}</math>, and <math>\overline{AM} \cdot \overline{MP}=\overline{XM} \cdot \overline{MC}</math> from before. This leads us to get that <math>\overline{BX} \cdot \overline {XQ}=\overline{XM} \cdot \overline{MC}</math>. | Now we start using Power of a Point. We get that <math>\overline{BX} \cdot \overline {XQ}= \overline{AM} \cdot \overline{MP}</math>, and <math>\overline{AM} \cdot \overline{MP}=\overline{XM} \cdot \overline{MC}</math> from before. This leads us to get that <math>\overline{BX} \cdot \overline {XQ}=\overline{XM} \cdot \overline{MC}</math>. | ||
− | Now we assign variables to the values of the segments. Let <math>\overline{BX}=a, \overline{XM}=b, \overline{MQ}=c, and \overline{QC}=d</math>. The equation from above gets us that <math>(a+b)c=b(c+d)</math>. As <math>a+b=c+d</math> from the problem statements, this gets us that <math>b=c</math> and <math>\overline{XC}=\overline{CQ}</math>, and we are done. | + | Now we assign variables to the values of the segments. Let <math>\overline{BX}=a, \overline{XM}=b, \overline{MQ}=c,</math> and <math>\overline{QC}=d</math>. The equation from above gets us that <math>(a+b)c=b(c+d)</math>. As <math>a+b=c+d</math> from the problem statements, this gets us that <math>b=c</math> and <math>\overline{XC}=\overline{CQ}</math>, and we are done. |
-dragoon and rhydon516 (: | -dragoon and rhydon516 (: | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>D</math> be the foot of the altitude from <math>A</math> onto <math>BC</math>. We want to show that <math>DM=MQ</math> for obvious reasons. | ||
+ | |||
+ | Notice that <math>ADPC</math> is cyclic and that <math>M</math> lies on the radical axis of <math>(ABPQ)</math> and <math>(ADPC)</math>. By Power of a Point, <math>(CM)(DM)=(BM)(MQ)</math>. As <math>BM=CM</math>, we have <math>DM=MQ</math>, as desired. | ||
+ | |||
+ | - Leo.Euler | ||
+ | |||
+ | ==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 | ||
+ | <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 | ||
+ | |||
+ | == 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> | ||
+ | |||
+ | ~ Daniel Ge | ||
+ | |||
+ | ==See Also== | ||
+ | {{USAJMO newbox|year=2023|num-b=1|num-a=3}} | ||
+ | {{MAA Notice}} |
Latest 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,
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.