## Task B-1.5.

Inside an equilateral triangle $ABC$ with side length $12 \, \text{cm}$, a point $T$ is chosen such that $|TP|:|TQ|:|TR|=1:2:3$, where $P, Q$, and $R$ are the feet of the perpendiculars from point $T$ to the sides of the triangle. What is the area of triangle $\triangle PQR$?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_aa38f3c63b8013e55583g-07.jpg?height=571&width=649&top_left_y=523&top_left_x=181)

Let $a=12 \text{ cm}$ be the side length of triangle $ABC$.

Denote $|TP|=k, |TQ|=2k, |TR|=3k$.

Notice that $|TP|, |TQ|$, and $|TR|$ are the heights of triangles $ABT, BCT$, and $CAT$, respectively.

Then for the areas of the triangles, we have:

$P_{ABC} = P_{ABT} + P_{BCT} + P_{CAT}$, or

$\frac{a^2 \sqrt{3}}{4} = \frac{a \cdot k}{2} + \frac{a \cdot 2k}{2} + \frac{a \cdot 3k}{2}$, from which it follows:

$\frac{12^2 \sqrt{3}}{4} = 6k + 12k + 18k$, so $k = \sqrt{3} \text{ cm}$.

Therefore, $|TP| = \sqrt{3} \text{ cm}, |TQ| = 2\sqrt{3} \text{ cm}$, and $|TR| = 3\sqrt{3} \text{ cm}$.

Notice that quadrilaterals $APTR, PBQT$, and $CQTR$ each have two right angles and one angle of $60^\circ$, and since the sum of the angles in a quadrilateral is $360^\circ$, we conclude that

$\angle PTQ = \angle QTR = \angle RTP = 120^\circ$.

The height of triangle $PQT$ is $\overline{DQ}$, which is also the leg of the right triangle $QDT$.

Then, $\sin 60^\circ = \frac{|DQ|}{2\sqrt{3}}$

or, $|DQ| = 3 \text{ cm}$.

Therefore, $P_{PQT} = \frac{\sqrt{3} \cdot 3}{2} = \frac{3\sqrt{3}}{2} \text{ cm}^2$.

Similarly, we obtain:

$P_{QRT} = \frac{9\sqrt{3}}{2} \text{ cm}^2$ and $P_{RPT} = \frac{9\sqrt{3}}{4} \text{ cm}^2$.

![](https://cdn.mathpix.com/cropped/2024_05_30_aa38f3c63b8013e55583g-07.jpg?height=506&width=779&top_left_y=1940&top_left_x=1027)

Then, $P(ABC) = P_{PQT} + P_{QRT} + P_{RPT} = \frac{3\sqrt{3}}{2} + \frac{9\sqrt{3}}{2} + \frac{9\sqrt{3}}{4} = \frac{33\sqrt{3}}{4} \text{ cm}^2$.

## NATIONAL MATHEMATICS COMPETITION

## 2nd grade - high school - B variant

## May 11, 2022.