## Task B-2.7.

Given is an isosceles triangle $A B C$ with base $\overline{A B}$ of length $14 \text{~cm}$ and legs of length $25 \text{~cm}$. Point $P$ is the midpoint of the base $\overline{A B}$. Point $Q$ is chosen on side $\overline{B C}$, and point $R$ is chosen on side $\overline{A C}$ such that $A B \| Q R$ and the area of triangle $P Q R$ is maximized. Determine the perimeter and area of triangle $P Q R$.
## Solution.

Let's sketch triangle $ABC$ and mark the given elements.

![](https://cdn.mathpix.com/cropped/2024_05_30_5ec97620b8cc23e4b297g-16.jpg?height=540&width=437&top_left_y=638&top_left_x=204)

1 point

Since $P$ is the midpoint of the base $\overline{AB}$, we know that the length of $\overline{CP}$ is the height of triangle $ABC$.

By applying the Pythagorean theorem to triangle $APC$, we can determine that

$|CP| = \sqrt{25^2 - 7^2} = \sqrt{(25-7)(25+7)} = \sqrt{18 \cdot 32} = \sqrt{9 \cdot 2 \cdot 2 \cdot 16} = 3 \cdot 2 \cdot 4 = 24 \text{ cm}$.

Let $S$ be the intersection of segments $\overline{CP}$ and $\overline{QR}$.

Let $x = |RS|$. We can observe that due to symmetry, $|SQ| = x$.

Since $AB \parallel QR$, we can conclude that triangles $APC$ and $RSC$ are similar by the $AA$ similarity criterion (one common angle, angles formed by the transversal $AC$ of the parallel lines $AB$ and $QR$).

Due to the similarity of the triangles, we know that $\frac{|RS|}{|AP|} = \frac{|CS|}{|CP|}$, or $\frac{x}{7} = \frac{|CS|}{24}$. 1 point We can express $|CS| = \frac{24}{7} x$.

The area of triangle $PQR$ is $P(PQR) = \frac{|QR| \cdot |PS|}{2}$.

We have $|QR| = 2x$ and $|PS| = |CP| - |CS| = 24 - \frac{24}{7} x$.

Thus, $P(PQR) = \frac{2x \cdot \left(24 - \frac{24}{7} x\right)}{2} = 24x - \frac{24}{7} x^2$.

Since the area of triangle $PQR$ is the maximum possible, with $a = -\frac{24}{7}$ and $b = 24$, we know that $x = -\frac{b}{2a} = -\frac{24}{2 \cdot \left(-\frac{24}{7}\right)} = \frac{7}{2} = 3.5 \text{ cm}$.

The area of triangle $PQR$ is $P(PQR) = 24 \cdot \frac{7}{2} - \frac{24}{7} \cdot \frac{49}{4} = 84 - 42 = 42 \text{ cm}^2$.

It remains to determine the perimeter of triangle $PQR$.

$o(PQR) = |PQ| + |QR| + |RP| = 2x + 2|PQ| = 7 + 2|PQ|$

To determine $|PQ|$, we apply the Pythagorean theorem to triangle $PQS$.

$|PS| = 24 - \frac{24}{7} x = 24 - \frac{24}{7} \cdot \frac{7}{2} = 24 - 12 = 12 \text{ cm}$

1 point

$|PQ| = \sqrt{\left(\frac{7}{2}\right)^2 + 12^2} = \sqrt{\frac{49}{4} + 144} = \sqrt{\frac{625}{4}} = \frac{25}{2} = 12.5 \text{ cm}$

1 point

Finally, $o(PQR) = 7 + 25 = 32 \text{ cm}$.

1 point

Note: The student can also determine the perimeter of triangle $PQR$ by concluding from the result $x = 3.5 \text{ cm}$ that $\overline{QR}$ is the midline of triangle $ABC$. From this conclusion, it follows that $\overline{PQ}$ and $\overline{PR}$ are also midlines of triangle $ABC$, so $|PQ| = |PR| = \frac{1}{2} \cdot 25 = 12.5 \text{ cm}$, or $o(PQR) = 7 + 2 \cdot 12.5 = 32 \text{ cm}$.

Note: If the student assumes, without proof, that the area of triangle $PQR$ is maximal when points $Q$ and $R$ are at the midpoints of their respective sides and solves the problem correctly, they can receive a maximum of 5 points.

## COUNTY COMPETITION IN MATHEMATICS

3rd grade - high school - B variant
March 4, 2023.

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