Task B-1.8. On the sides $\overline{B C}$ and $\overline{C D}$ of the square $A B C D$, points $P$ and $Q$ are given such that

$\overline{B P}: \overline{P C}=\overline{C Q}: \overline{Q D}=1: 3$. If the length of the side of the square is $a$, determine the length of the greatest and the smallest height of the triangle $A P Q$.
Solution. Sketch

![](https://cdn.mathpix.com/cropped/2024_05_30_cf2c4989ba380bc8684dg-04.jpg?height=734&width=683&top_left_y=625&top_left_x=688)

Let the sides of triangle $A P Q$ be denoted as $x, y, z$. We calculate their lengths using the Pythagorean theorem.

$$
\begin{array}{r}
x^{2}=a^{2}+\frac{1}{16} a^{2}, \quad x=\frac{\sqrt{17}}{4} a \\
y^{2}=\frac{9}{16} a^{2}+\frac{1}{16} a^{2}=\frac{10}{16} a^{2}, \quad y=\frac{\sqrt{10}}{4} a \\
z^{2}=a^{2}+\frac{9}{16} a^{2}=\frac{25}{16} a^{2}, \quad z=\frac{5}{4} a
\end{array}
$$

(Note: it is sufficient to calculate two sides $y$ and $z$)

The longest height is to the shortest side, and the shortest height is to the longest side. (1 point) We will calculate them using the area of triangle $A P Q$.

$$
P=a^{2}-\frac{1}{2} \cdot \frac{1}{4} a \cdot a-\frac{1}{2} \cdot \frac{3}{4} a \cdot \frac{1}{4} a-\frac{1}{2} \cdot \frac{3}{4} a \cdot a=\frac{13}{32} a^{2}
$$

Then the shortest height is

$$
v_{z}=\frac{2 P}{z}=\frac{2 \cdot \frac{13}{32} a^{2}}{\frac{5}{4} a}=\frac{13}{20} a
$$

The longest height is

$$
v_{y}=\frac{2 P}{y}=\frac{2 \cdot \frac{13}{32} a^{2}}{\frac{\sqrt{10}}{4} a}=\frac{13}{4 \sqrt{10}} a=\frac{13 \sqrt{10}}{40} a
$$

## SCHOOL/CITY COMPETITION IN MATHEMATICS

## 2nd grade - high school - B variant

14th February 2012.

IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO EVALUATE THAT APPROACH APPROPRIATELY.