Subfield: Geometry


Question: Points $A$ and $L$ lie outside circle $\omega$, whose center is $O$, and $\overline{A L}$ contains diameter $\overline{R M}$, as shown below. Circle $\omega$ is tangent to $\overline{L K}$ at $K$. Also, $\overline{A K}$ intersects $\omega$ at $Y$, which is between $A$ and $K$. If $K L=3, M L=2$, and $\mathrm{m} \angle A K L-\mathrm{m} \angle Y M K=90^{\circ}$, compute $[A K M]$ (i.e., the area of $\triangle A K M$ ).

![](https://cdn.mathpix.com/cropped/2023_12_21_32b17155940ff395abf0g-1.jpg?height=396&width=740&top_left_y=1057&top_left_x=736)

Solution:
Notice that $\overline{O K} \perp \overline{K L}$, and let $r$ be the radius of $\omega$.

![](https://cdn.mathpix.com/cropped/2023_12_21_bd4901b10dbf246bfe20g-1.jpg?height=404&width=742&top_left_y=351&top_left_x=735)

Then consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\frac{5}{4}$.

Because $\mathrm{m} \angle Y K L=\frac{1}{2} \mathrm{~m} \widehat{Y R K}$ and $\mathrm{m} \angle Y M K=\frac{1}{2} \mathrm{~m} \widehat{Y K}$, it follows that $\mathrm{m} \angle Y K L+\mathrm{m} \angle Y M K=$ $180^{\circ}$. By the given condition, $\mathrm{m} \angle Y K L-\mathrm{m} \angle Y M K=90^{\circ}$. It follows that $\mathrm{m} \angle Y M K=45^{\circ}$ and $\mathrm{m} \angle Y K L=135^{\circ}$. hence $\mathrm{m} \widehat{Y K}=90^{\circ}$. Thus,

$$
\overline{Y O} \perp \overline{O K} \quad \text { and } \quad \overline{Y O} \| \overline{K L}
\tag{*}
$$
Compute $[A K M]$ as $\frac{1}{2}$ base $\cdot$ height, using base $\overline{A M}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_bd4901b10dbf246bfe20g-1.jpg?height=407&width=746&top_left_y=1428&top_left_x=733)

Because of (*), $\triangle A Y O \sim \triangle A K L$. To compute $A M$, notice that in $\triangle A Y O, A O=A M-r$, while in $\triangle A K L$, the corresponding side $A L=A M+M L=A M+2$. Therefore:

$$
\begin{aligned}
\frac{A O}{A L} & =\frac{Y O}{K L} \\
\frac{A M-\frac{5}{4}}{A M+2} & =\frac{5 / 4}{3}
\end{aligned}
$$

from which $A M=\frac{25}{7}$. Draw the altitude of $\triangle A K M$ from vertex $K$, and let $h$ be its length. In right triangle $O K L, h$ is the altitude to the hypotenuse, so $\frac{h}{3}=\sin (\angle K L O)=\frac{r}{r+2}$. Hence $h=\frac{15}{13}$. Therefore $[A K M]=\frac{1}{2} \cdot \frac{25}{7} \cdot \frac{15}{13}=\frac{375}{182}$.
Notice that $\overline{O K} \perp \overline{K L}$, and let $r$ be the radius of $\omega$.

![](https://cdn.mathpix.com/cropped/2023_12_21_bd4901b10dbf246bfe20g-1.jpg?height=404&width=742&top_left_y=351&top_left_x=735)

Then consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\frac{5}{4}$.

Because $\mathrm{m} \angle Y K L=\frac{1}{2} \mathrm{~m} \widehat{Y R K}$ and $\mathrm{m} \angle Y M K=\frac{1}{2} \mathrm{~m} \widehat{Y K}$, it follows that $\mathrm{m} \angle Y K L+\mathrm{m} \angle Y M K=$ $180^{\circ}$. By the given condition, $\mathrm{m} \angle Y K L-\mathrm{m} \angle Y M K=90^{\circ}$. It follows that $\mathrm{m} \angle Y M K=45^{\circ}$ and $\mathrm{m} \angle Y K L=135^{\circ}$. hence $\mathrm{m} \widehat{Y K}=90^{\circ}$. Thus,

$$
\overline{Y O} \perp \overline{O K} \quad \text { and } \quad \overline{Y O} \| \overline{K L}
\tag{*}
$$
By the Power of the Point Theorem, $L K^{2}=L M \cdot L R$, so

$$
\begin{aligned}
L R & =\frac{9}{2} \\
R M & =L R-L M=\frac{5}{2} \\
O L & =r+M L=\frac{13}{4}
\end{aligned}
$$

From (*), we know that $\triangle A Y O \sim \triangle A K L$. Hence by $(\dagger)$,

$$
\frac{A L}{A O}=\frac{A L}{A L-O L}=\frac{K L}{Y O}=\frac{3}{5 / 4}=\frac{12}{5}, \quad \text { thus } \quad A L=\frac{12}{7} \cdot O L=\frac{12}{7} \cdot \frac{13}{4}=\frac{39}{7}
$$

Hence $A M=A L-2=\frac{25}{7}$. The ratio between the areas of triangles $A K M$ and $R K M$ is equal to

$$
\frac{[A K M]}{[R K M]}=\frac{A M}{R M}=\frac{25 / 7}{5 / 2}=\frac{10}{7}
$$

Thus $[A K M]=\frac{10}{7} \cdot[R K M]$.

Because $\angle K R L$ and $\angle M K L$ both subtend $\widehat{K M}, \triangle K R L \sim \triangle M K L$. Therefore $\frac{K R}{M K}=\frac{L K}{L M}=$ $\frac{3}{2}$. Thus let $K R=3 x$ and $M K=2 x$ for some positive real number $x$. Because $R M$ is a diameter of $\omega$ (see left diagram below), $\mathrm{m} \angle R K M=90^{\circ}$. Thus triangle $R K M$ is a right triangle with hypotenuse $\overline{R M}$. In particular, $13 x^{2}=K R^{2}+M K^{2}=R M^{2}=\frac{25}{4}$, so $x^{2}=\frac{25}{52}$ and $[R K M]=\frac{R K \cdot K M}{2}=3 x^{2}$. Therefore

$$
[A K M]=\frac{10}{7} \cdot[R K M]=\frac{10}{7} \cdot 3 \cdot \frac{25}{52}=\frac{\mathbf{3 7 5}}{\mathbf{1 8 2}}
$$

![](https://cdn.mathpix.com/cropped/2023_12_21_352f1bc0cb3b408f3284g-1.jpg?height=364&width=1484&top_left_y=1650&top_left_x=367)
Notice that $\overline{O K} \perp \overline{K L}$, and let $r$ be the radius of $\omega$.

![](https://cdn.mathpix.com/cropped/2023_12_21_bd4901b10dbf246bfe20g-1.jpg?height=404&width=742&top_left_y=351&top_left_x=735)

Then consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\frac{5}{4}$.

Because $\mathrm{m} \angle Y K L=\frac{1}{2} \mathrm{~m} \widehat{Y R K}$ and $\mathrm{m} \angle Y M K=\frac{1}{2} \mathrm{~m} \widehat{Y K}$, it follows that $\mathrm{m} \angle Y K L+\mathrm{m} \angle Y M K=$ $180^{\circ}$. By the given condition, $\mathrm{m} \angle Y K L-\mathrm{m} \angle Y M K=90^{\circ}$. It follows that $\mathrm{m} \angle Y M K=45^{\circ}$ and $\mathrm{m} \angle Y K L=135^{\circ}$. hence $\mathrm{m} \widehat{Y K}=90^{\circ}$. Thus,

$$
\overline{Y O} \perp \overline{O K} \quad \text { and } \quad \overline{Y O} \| \overline{K L}
\tag{*}
$$

Let $U$ and $V$ be the respective feet of the perpendiculars dropped from $A$ and $M$ to $\overleftrightarrow{K L}$. From (*), $\triangle A K L$ can be dissected into two infinite progressions of triangles: one progression of triangles similar to $\triangle O K L$ and the other similar to $\triangle Y O K$, as shown in the right diagram above. In both progressions, the corresponding sides of the triangles have common ratio equal to

$$
\frac{Y O}{K L}=\frac{5 / 4}{3}=\frac{5}{12}
$$



Thus

$$
A U=\frac{5}{4}\left(1+\frac{5}{12}+\left(\frac{5}{12}\right)^{2}+\cdots\right)=\frac{5}{4} \cdot \frac{12}{7}=\frac{15}{7}
$$

Because $\triangle L M V \sim \triangle L O K$, and because $L O=\frac{13}{4}$ by $(\dagger)$,

$$
\frac{M V}{O K}=\frac{L M}{L O}, \quad \text { thus } \quad M V=\frac{O K \cdot L M}{L O}=\frac{\frac{5}{4} \cdot 2}{\frac{13}{4}}=\frac{10}{13}
$$

Finally, note that $[A K M]=[A K L]-[K L M]$. Because $\triangle A K L$ and $\triangle K L M$ share base $\overline{K L}$,

$$
[A K M]=\frac{1}{2} \cdot 3 \cdot\left(\frac{15}{7}-\frac{10}{13}\right)=\frac{\mathbf{3 7 5}}{\mathbf{1 8 2}}
$$

Final answer: $\frac{375}{182}$