Inside triangle $A B C$, a point $K$ is taken. It is known that $A K=1, K C=\sqrt{3}$, and the angles $A K C, A B K$, and $K B C$ are $120^{\circ}$, $15^{\circ}$, and $15^{\circ}$ respectively. Find $B K$.
Let $P$ and $Q$ be the projections of point $K$ onto sides $AB$ and $BC$. Consider triangles $APK$ and $KQC$.

## Solution

Let $P$ and $Q$ be the feet of the perpendiculars dropped from point $K$ to sides $AB$ and $BC$ respectively. Then $\angle AKP + \angle CKQ = 360^\circ - \angle AKC - \angle PKQ = 360^\circ - 120^\circ - 150^\circ = 90^\circ$. Let $\angle PAK = \alpha$. Then $\angle CKQ = 90^\circ - \angle AKP = \alpha$; $KQ = KC \cos \alpha = \sqrt{3} \cos \alpha$,

$KP = AK \sin \alpha = \sin \alpha$, and since $KQ = KP$, then $\tan \alpha = \sqrt{3}, \alpha = 60^\circ, KP = AK \sin 60^\circ = \frac{\sqrt{3}}{2}$. Since $\sin 15^\circ = \sqrt{\frac{1 - \cos 30^\circ}{2}} = \sqrt{\frac{4 - 2\sqrt{3}}{8}} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$, then

$BK = \frac{KP}{\sin \angle PKB} = \frac{\sqrt{6}}{\sqrt{3} - 1} = \frac{\sqrt{6}(\sqrt{3} + 1)}{2} = \frac{3\sqrt{2} + \sqrt{6}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_cd0469c76668f3ff0930g-07.jpg?height=318&width=326&top_left_y=2585&top_left_x=869)

## [ Angle Bisector (gMT). $\quad]$  Problem 54456 Topics: [ Ratio in which the angle bisector divides the side.] Heron's Formula

On the bisector of the acute angle $AOC$, a point $B$ is taken. Through point $B$, a line perpendicular to $OB$ is drawn, intersecting side $AO$ at point $K$ and side $OC$ at point $L$. Through point $B$, another line is drawn, intersecting side $AO$ at point $M$ (with $M$ between $O$ and $K$), and side $OC$ at point $N$, such that $\angle MON = \angle MNO$. It is known that $MK = a$, $LN = 3a / 2$. Find the area of triangle $MON$.

## Solution

Consider point $P$, which is symmetric to $N$ with respect to line $OB$. This point lies on line $OM$, and $KP = LN$. Moreover, $\angle KBP = \angle LBN = \angle KBM$, so $BK$ is the bisector of angle $MBP$. Therefore, $2: 3 = MK: LN = MK: KP = MB: PB = MB: BN = MO: NO = OM: OP$. Consequently, $OM = 2MP = 2(a + 3a / 2) = 5a, ON = 3 / 2 OM = 7.5a$.

![](https://cdn.mathpix.com/cropped/2024_05_06_cd0469c76668f3ff0930g-08.jpg?height=580&width=829&top_left_y=1002&top_left_x=617)

The area of triangle $MON$ with sides $5a, 5a, 7.5a$ can be found using Heron's formula:

$S = \frac{1}{4} \sqrt{17.5a \cdot 7.5a \cdot 7.5a \cdot 2.5a} = \frac{75a^2 \sqrt{7}}{16}$.

## Answer

$75a^2 \sqrt{7}$

16