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

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==Problem==
 
==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>.
 
(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>.
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==Video Solution==
 +
https://www.youtube.com/watch?v=f-d4mi-AyxQ
  
 
==Solution 1==
 
==Solution 1==

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

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

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