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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 (:

Revision as of 15:34, 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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