4. Let $\triangle ABC$ be an isosceles triangle with legs of length $2 \sqrt{2} \mathrm{~cm}$. Let $P$ be the midpoint of the height dropped to the base of this triangle. If the distance from point $P$ to the leg is three times smaller than its distance to the base, what is the area of this triangle?
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_50030aca1ecc6c9ce9eeg-2.jpg?height=685&width=485&top_left_y=1916&top_left_x=794)

Let point $D$ be the foot of the altitude from the base of the triangle. Point $P$ is the midpoint of the altitude $\overline{A D}$.

Let points $E$ and $F$ be the feet of the perpendiculars from point $P$ to the sides $\overline{A B}$ and $\overline{A C}$.

Let $|B C|=a$ and $|A B|=|A C|=b=2 \sqrt{2} \text{ cm}$.

Triangles $AFP$ and $APE$ are congruent by the SAS (Side-Angle-Side) congruence theorem ($\overline{A P}$ is a common side, $|\angle P A F|=|\angle P A E|$ because triangle $A B C$ is isosceles, $|\angle P F A|=|\angle A E P|=90^{\circ}$, so $|\angle A P F|=|\angle E P A|$).

Let $|P E|=|P F|=k, k>0$. From the given conditions, it follows that $|P D|=3 k$, so the length of the altitude to the base is $v_{a}=|A D|=6 k$.

1 POINT

The area of triangle $A B C$ is equal to the sum of the areas of triangles $B C P$, $C A P$, and $A B P$. We have:

$p_{\triangle A B C}=p_{\triangle B C P}+p_{\triangle C A P}+p_{\triangle A B P}$

$\frac{|B C| \cdot|A D|}{2}=\frac{|B C| \cdot|P D|}{2}+\frac{|A C| \cdot|P E|}{2}+\frac{|A B| \cdot|P F|}{2}$

2 POINTS

$\frac{a \cdot 6 k}{2}=\frac{a \cdot 3 k}{2}+\frac{b \cdot k}{2}+\frac{b \cdot k}{2} \quad / \cdot 2$

$6 a k=3 a k+2 b k$

$3 a=2 b$

1 POINT

$a=\frac{2}{3} \cdot b=\frac{2}{3} \cdot 2 \sqrt{2}=\frac{4}{3} \sqrt{2} \text{ cm}$

1 POINT

By applying the Pythagorean theorem to triangle $A D C$, we calculate the length of the altitude:

$|A D|^{2}=|A C|^{2}-|D C|^{2}$

$v_{a}^{2}=b^{2}-\left(\frac{1}{2} \cdot a\right)^{2}$

$v_{a}^{2}=(2 \sqrt{2})^{2}-\left(\frac{1}{2} \cdot \frac{4}{3} \sqrt{2}\right)^{2}$

1 POINT

$v_{a}^{2}=8-\frac{8}{9}$

$v_{a}^{2}=\frac{64}{9}$

$v_{a}=\frac{8}{3} \text{ cm}$

1 POINT

Finally, $P=\frac{a \cdot v_{a}}{2}=\frac{1}{2} \cdot \frac{4}{3} \sqrt{2} \cdot \frac{8}{3}=\frac{16}{9} \sqrt{2} \text{ cm}^{2}$.

1 POINT