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In triangle PQR, M is the midpoint of \overline{PQ}. Let X be the point on \overline{QR} such that \overline{PX} bisects \angle{QPR} and let the perpendicular bisector of \overline{PQ} intersect \overline{PX} at Y. If PQ = 36, PR = 22, and MY = 8, then find the area of triangle PYR.

 Apr 30, 2024
 #1
avatar+9664 
+2

Note that PX bisects \(\angle QPR\). Then, we can suppose \(\angle XPQ = \angle XPR = \theta\).

Recall the formula:

If two sides of a triangle are of length a and b respectively, and the angle included by these two sides is \(\phi\), then the area of the triangle is \(\dfrac12 ab \sin \phi\).

Now, draw the diagram and look at triangle PMY. Its height is MY (which is 8) and its base is PM (which is 18). So the area of triangle PMY is \(\dfrac12 \cdot 8 \cdot 18 = 72\). But on the other hand, using the formula stated above, the same area is also equal to \(\dfrac12 (PM)(PY) \sin \theta = 9\cdot PY \sin \theta\).

Therefore, we know that \(PY \sin \theta = 8\) by comparing.

 

Now, we look at triangle PYR. By the formula, the area is \(\dfrac12 (PY)(PR) \sin \theta\), but we know PR = 22. Using \(PY \sin \theta = 8\), the required area is \(\dfrac12 (22) (PY \sin \theta) = 11(8) = \boxed{88}\).

 Apr 30, 2024
 #2
avatar+42 
+2

I managed to solve it myself and got the same thing!!

 Apr 30, 2024

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