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

(Created page with "==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 perpend...")
 
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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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==Solution 1==
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The condition is solved only if <math>\triangle{NBC}</math> is isosceles, which in turn only happens if <math>\overline{MN}</math> is perpendicular to <math>\overline{BC}</math>.
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Now, draw the altitude from <math>A</math> to <math>\overline{BC}</math>, and call that point <math>X</math>. Because of the Midline Theorem, the only way that this condition is met is if <math>\triangle{AXQ} \sim \triangle{NMQ}</math>, or if <math>\overline{XM}=\overline{MQ}</math>.
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By <math>AA</math> similarity, <math>\triangle{AXM} \sim \triangle{CPM}</math>. Using similarity ratios, we get that <math>\frac{\overline{AM}}{\overline{XM}}=\frac{\overline{CM}}{\overline{PM}}</math>. Rearranging, we get that <math>\overline{AM} \cdot \overline{MP}=\overline{XM} \cdot \overline{MC}</math>. This implies that <math>AXPC</math> is cyclic.
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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>.
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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.
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-dragoon and rhydon516

Revision as of 14:11, 24 March 2023

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 $NB=NC$.

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

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