4. In a right-angled triangle where the legs are in the ratio 5:12, the circle with diameter $MN$ where $M$ and $N$ are the midpoints of the legs, intersects the hypotenuse at points $P$ and $Q$. Determine the ratio in which points $P$ and $Q$ divide the hypotenuse.
Solution. Let the legs of the triangle $ABC$ satisfy $\overline{AC}: \overline{BC}=5: 12$, so $\overline{AC}=5k$ and $\overline{BC}=12k$, for some $k>0$ (see the diagram on the right). Then

$$
\overline{AB}=\sqrt{\overline{AC}^{2}+\overline{BC}^{2}}=\sqrt{25k^{2}+144k^{2}}=13k
$$

Since $MN$ is the midline of the triangle $ABC$,

![](https://cdn.mathpix.com/cropped/2024_06_05_31995039ed3f7c75b5c4g-02.jpg?height=271&width=466&top_left_y=1119&top_left_x=995)
we have $\overline{MN}=\frac{1}{2} \overline{AB}=\frac{13}{2} k$. The circle with diameter $MN$ passes through the vertex $C$, because the angle at vertex $C$ is a right angle. The point that is symmetric to point $C$ with respect to the diameter $MN$ is also a point on the circle with diameter $MN$. Let this point be $P'$. From $CP' \perp MN$ and $MN \| AB$ it follows that $CP' \perp AB$. Since $MN$ is the midline of the triangle $ABC$, we conclude that the point $P'$ lies on the side $AB$. Therefore, $P' \equiv P$ and $CP$ is the altitude of the triangle $ABC$ drawn to the hypotenuse. For the area of the triangle, we have $\frac{\overline{AC} \cdot \overline{BC}}{2}=\frac{\overline{AB} \cdot \overline{CP}}{2}$, from which we find $\frac{5k \cdot 12k}{2}=\frac{13k \cdot \overline{CP}}{2}$, or $\overline{CP}=\frac{60}{13} k$. Then, from the right triangle $APC$ we have $\overline{AP}=\sqrt{\overline{AC}^{2}-\overline{CP}^{2}}=\sqrt{25k^{2}-\frac{3600}{169}k^{2}}=\frac{25}{13} k$.

For the second intersection point $Q$, from $\measuredangle QPC=90^{\circ}$, it follows that $CQ$ is the diameter of the circle, so $\overline{CQ}=\frac{13}{2} k$. Now, from the right triangle $QPC$ we have

$$
\overline{PQ}=\sqrt{\overline{CQ}^{2}-\overline{CP}^{2}}=\sqrt{\frac{169}{4}k^{2}-\frac{3600}{169}k^{2}}=\frac{119}{26} k
$$

Then,

$$
\overline{QB}=\overline{AB}-\overline{AP}-\overline{PQ}=13k-\frac{25}{13}k-\frac{119}{26}k=\frac{13}{2}k
$$

Finally, $\overline{AP}: \overline{PQ}: \overline{QB}=\frac{25}{13}k: \frac{119}{26}k: \frac{13}{2}k=50: 119: 169$.

## 2nd Year