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# Geometry

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In triangle $PQR$, let $M$ be the midpoint of $\overline{PQ}$, let $N$ be the midpoint of $\overline{PR}$, and let $O$ be the intersection of $\overline{QN}$ and $\overline{RM}$, as shown. If $\overline{QN}\perp\overline{PR}$, $QN = 15$, and $PR = 20$, then find $QR$.

Apr 26, 2024

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In triangle $PQR$, we are given that $M$ is the midpoint of $\overline{PQ}$, $N$ is the midpoint of $\overline{PR}$, and $O$ is the intersection of $\overline{QN}$ and $\overline{RM}$. Additionally, we are told that $\overline{QN}\perp\overline{PR}$, $QN = 15$, and $PR = 20$.

Since $M$ is the midpoint of $\overline{PQ}$ and $N$ is the midpoint of $\overline{PR}$, we can conclude that $MO$ is the median of triangle $PQR$. Therefore, $MO$ passes through the midpoint of $\overline{QR}$.

Since $\overline{QN}\perp\overline{PR}$, triangle $PQN$ is a right triangle. According to the Pythagorean theorem, we have:

ππ2=ππ2+ππ2PQ2=QN2+PR2 ππ2=152+202PQ2=152+202 ππ2=225+400PQ2=225+400 ππ2=625PQ2=625 ππ=25PQ=25

Since $M$ is the midpoint of $\overline{PQ}$, we have $PM = MQ = \frac{1}{2}PQ = \frac{1}{2} \times 25 = 12.5$.

Now, since $MO$ is the median of triangle $PQR$, we have $MO = \frac{1}{2}QR$. Therefore, $QR = 2 \times MO$.

To find $MO$, we can use the fact that $MO$ is perpendicular to $QN$, so triangles $QMO$ and $QNO$ are similar triangles. Therefore:

ππππ=ππππQNMOβ=QRQNβ ππ15=15ππ15MOβ=QR15β ππ=152ππMO=QR152β ππ=225ππMO=QR225β

Since $MO = \frac{1}{2}QR$, we can set up the following equation:

12ππ=225ππ21βQR=QR225β

Solving for $QR$, we get:

ππ2=450QR2=450

ππ=450QR=450β

ππ=1510QR=1510β

Therefore, the length of $QR$ is $15\sqrt{10}$.

Apr 27, 2024