Difference between revisions of "2023 USAJMO Problems/Problem 2"

Latest revision as of 15:16, 15 September 2024

Problem

(Holden Mui) In an acute triangle $ABC$ , let $M$ be the midpoint of $\overline{BC}$ . Let $P$ be the foot of the perpendicular from $C$ to $AM$ . Suppose that the circumcircle of triangle $ABP$ intersects line $BC$ at two distinct points $B$ and $Q$ . Let $N$ be the midpoint of $\overline{AQ}$ . Prove that $NB=NC$ .

Video Solution

https://www.youtube.com/watch?v=f-d4mi-AyxQ

Solution 1

The condition is solved only if $\triangle{NBC}$ is isosceles, which in turn only happens if $\overline{MN}$ is perpendicular to $\overline{BC}$ .

Now, draw the altitude from $A$ to $\overline{BC}$ , and call that point $X$ . Because of the Midline Theorem, the only way that this condition is met is if $\triangle{AXQ} \sim \triangle{NMQ}$ , or if $\overline{XM}=\overline{MQ}$ .

By $AA$ similarity, $\triangle{AXM} \sim \triangle{CPM}$ . Using similarity ratios, we get that $\frac{\overline{AM}}{\overline{XM}}=\frac{\overline{CM}}{\overline{PM}}$ . Rearranging, we get that $\overline{AM} \cdot \overline{MP}=\overline{XM} \cdot \overline{MC}$ . This implies that $AXPC$ is cyclic.

Now we start using Power of a Point. We get that $\overline{BX} \cdot \overline {XQ}= \overline{AM} \cdot \overline{MP}$ , and $\overline{AM} \cdot \overline{MP}=\overline{XM} \cdot \overline{MC}$ from before. This leads us to get that $\overline{BX} \cdot \overline {XQ}=\overline{XM} \cdot \overline{MC}$ .

Now we assign variables to the values of the segments. Let $\overline{BX}=a, \overline{XM}=b, \overline{MQ}=c,$ and $\overline{QC}=d$ . The equation from above gets us that $(a+b)c=b(c+d)$ . As $a+b=c+d$ from the problem statements, this gets us that $b=c$ and $\overline{XC}=\overline{CQ}$ , and we are done.

-dragoon and rhydon516 (:

Solution 2

Let $D$ be the foot of the altitude from $A$ onto $BC$ . We want to show that $DM=MQ$ for obvious reasons.

Notice that $ADPC$ is cyclic and that $M$ lies on the radical axis of $(ABPQ)$ and $(ADPC)$ . By Power of a Point, $(CM)(DM)=(BM)(MQ)$ . As $BM=CM$ , we have $DM=MQ$ , as desired.

- Leo.Euler

Solution 3 (Less technical bary)

We are going to use barycentric coordinates on $\triangle ABC$ . Let $A=(1,0,0)$ , $B=(0,1,0)$ , $C=(0,0,1)$ , and $a=BC$ , $b=CA$ , $c=AB$ . We have $M=\left(0,\frac{1}{2},\frac{1}{2}\right)$ and $P=(x:1:1)$ so $\overrightarrow{CP}=\left(\frac{x}{x+2},\frac{1}{x+2},\frac{1}{x+2}-1\right)$ and $\overrightarrow{AM}=\left(-1,\frac{1}{2},\frac{1}{2}\right)$ . Since $\overleftrightarrow{CP}\perp\overleftrightarrow{AM}$ , 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 $(ABP)$ is $-a^2yz-b^2zx-c^2xy+ux+vy+wz=0.$ Plugging in $A$ and $B$ gives $u=v=0$ . Plugging in $P$ 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 $Q=(0,t,1-t)$ where $\begin{align*} -a^2t(1-t)+w(1-t)&=0\\ \implies t&=\frac{w}{a^2} \end{align*}$ so $Q=\left(0,\frac{w}{a^2},1-\frac{w}{a^2}\right)$ . It follows that $N=\left(\frac{1}{2},\frac{w}{2a^2},1-\frac{w}{2a^2}\right)$ . It suffices to prove that $\overleftrightarrow{ON}\perp\overleftrightarrow{BC}$ . Setting $\overrightarrow{O}=0$ , we get $\overrightarrow{N}=\left(\frac{1}{2},\frac{w}{2a^2},1-\frac{w}{2a^2}\right)$ . Furthermore we have $\overrightarrow{CB}=(0,1,-1)$ 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. $\square$

~KevinYang2.71

Solution 4 (Less bashy bary)

We employ barycentric coordinates. Set $AMC$ as the reference triangle with $A = (1, 0, 0)$ , $M = (0, 1, 0)$ , and $C = (0, 0, 1)$ . We immediately have, $P = (S_B : S_A : 0); B = (0, 2, -1)$ Since it passes through $A$ , for some $v, w$ , the equation of circle $(ABP)$ is, $(ABP): -a^2 yz - b^2 zx - c^2 xy + (vy + wz)(x + y + z) = 0$ Plugging in $P$ , $- c^2 S_{AB} + (v S_A)(c^2) = 0$ $\iff v = S_B$ Plugging in $B$ , $2a^2 + (2v - w) = 0$ $\iff 2a^2 + 2 S_B = 3 a^2 - b^2 + c^2 = w$ In conclusion the circle has formula, $(ABP): -a^2 yz - b^2 zx - c^2 xy + ((S_B)y + (3 a^2 - b^2 + c^2) z)(x + y + z) = 0$ $Q$ is the second intersection of circle $(ABP)$ with $\overline{CM}$ . We let $Q = (0, 1 - t, t)$ for some $t \neq -1$ . Plugging this in, $-a^2 (1-t)t + ((S_B)(1-t) + (3 a^2 - b^2 + c^2) t) = 0$ We claim that $t = -\frac{S_B}{a^2}$ is the other solution. $\left(1+ \frac{S_B}{a^2} \right) S_B + \left((S_B)\left(1+ \frac{S_B}{a^2}\right) - (3 a^2 - b^2 + c^2) \frac{S_B}{a^2}\right) = 0$ $\iff \left(\frac{3a^2 - b^2 + c^2}{2a^2} \right) + \left(\left(\frac{3a^2 - b^2 + c^2}{2a^2}\right) - (3 a^2 - b^2 + c^2) \frac{1}{a^2}\right) = 0$ Factoring out the $\frac{3a^2 - b^2 + c^2}{2a^2}$ , this is clearly true.

We also check that, these are not the same value. $-\frac{S_B}{a^2} = -1$ $\iff a^2 - b^2 + c^2 = 2a^2$ $\iff c^2 = a^2 + b^2$ The triangle is acute, so this is impossible.

Since we had a quadratic in $t$ with at most two solutions, the second intersection $Q$ is indeed, $Q = \left( 0, 1 + \frac{S_B}{a^2}, -\frac{S_B}{a^2} \right)$ Therefore, $N = \frac{A + Q}{2} = \left( \frac{1}{2}, \frac{1}{2} - \frac{S_B}{2a^2}, \frac{S_B}{2a^2} \right)$

~ Daniel Ge

@@ Line 1: / Line 1: @@
 ==Problem==
 (Holden Mui) In an acute triangle <math>ABC</math>, let <math>M</math> be the midpoint of <math>\overline{BC}</math>. Let <math>P</math> be the foot of the perpendicular from <math>C</math> to <math>AM</math>. Suppose that the circumcircle of triangle <math>ABP</math> intersects line <math>BC</math> at two distinct points <math>B</math> and <math>Q</math>. Let <math>N</math> be the midpoint of <math>\overline{AQ}</math>. Prove that <math>NB=NC</math>.
+==Video Solution==
+https://www.youtube.com/watch?v=f-d4mi-AyxQ
 ==Solution 1==
@@ Line 23: / Line 26: @@
 - 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}}

2023 USAJMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
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All USAJMO Problems and Solutions

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