4. The height to the base of an isosceles triangle has a length of $\sqrt{2+\sqrt{2}}$. If the measure of the angle opposite the base is $45^{\circ}$, what is the length of the height to the leg of the triangle?
4. The height to the base of an isosceles triangle has a length of $\sqrt{2+\sqrt{2}}$. If the measure of the angle opposite the base is $45^{\circ}$, what is the length of the height to the leg of this triangle?

## Solution.

Let $\overline{A B}$ be the base of the isosceles triangle $\triangle A B C$ and $|\angle A C B|=45^{\circ}$.

Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_ca549eb60a6f3c3b6299g-32.jpg?height=863&width=737&top_left_y=1436&top_left_x=268)

Let $D$ be the foot of the altitude from vertex $C$. Then $v=|C D|=\sqrt{2+\sqrt{2}}$.

Let $E$ be the foot of the altitude from vertex $A$ to side $\overline{B C}$.

Triangle $\triangle A E C$ is a right triangle, and since $|\angle A C B|=45^{\circ}$, it follows that $|\angle E A C|=180^{\circ}-\left(90^{\circ}+45^{\circ}\right)=45^{\circ}$.

1 POINT

Thus, triangle $\triangle A E C$ is isosceles, so $|A E|=|E C|=x$.

Furthermore, by the Pythagorean theorem, $|A C|=\sqrt{x^{2}+x^{2}}=\sqrt{2 x^{2}}=x \sqrt{2}$.

1 POINT

Let $|B E|=y$.

Since $\triangle A B C$ is isosceles, it follows that $x+y=x \sqrt{2}$.

From this, we get $y=x \sqrt{2}-x=x(\sqrt{2}-1)$.

Triangle $\triangle A B E$ is also a right triangle, and it holds that $|A B|^{2}=x^{2}+y^{2}$ (Pythagorean theorem).

On the other hand,

$$
\begin{aligned}
x^{2}+y^{2} & =x^{2}+x^{2}(\sqrt{2}-1)^{2} \\
& =x^{2}+x^{2}(2-2 \sqrt{2}+1) \\
& =x^{2}+x^{2}(3-2 \sqrt{2}) \\
& =x^{2}(1+3-2 \sqrt{2}) \\
& =x^{2}(4-2 \sqrt{2})
\end{aligned}
$$

so $|A B|=\sqrt{x^{2}+y^{2}}=\sqrt{x^{2}(4-2 \sqrt{2})}=x \sqrt{4-2 \sqrt{2}}=x \sqrt{2} \cdot \sqrt{2-\sqrt{2}}$.

The area of triangle $\triangle A B C$ is

$P=\frac{1}{2}|A B| \cdot v=\frac{1}{2}|B C| \cdot x$.

From this, it follows that

$|A B| \cdot v=|B C| \cdot x$

$x \sqrt{2} \cdot \sqrt{2-\sqrt{2}} \cdot \sqrt{2+\sqrt{2}}=x \sqrt{2} \cdot x /: x \sqrt{2}$

$\sqrt{2-\sqrt{2}} \cdot \sqrt{2+\sqrt{2}}=x$

$x=\sqrt{(2-\sqrt{2}) \cdot(2+\sqrt{2})}$

$x=\sqrt{4-2}$

The length of the height to the leg is $x=\sqrt{2}$.

Note 1: After calculating $|A B|$, the student can obtain $x$ by applying the Pythagorean theorem. In the right triangle $\triangle A D C$,

$$
|A C|^{2}=v^{2}+\left(\frac{|A B|}{2}\right)^{2}
$$

or

$$
(x \sqrt{2})^{2}=(\sqrt{2+\sqrt{2}})^{2}+\left(\frac{x \sqrt{2} \cdot \sqrt{2-\sqrt{2}}}{2}\right)^{2}
$$

Further,

$2 x^{2}=2+\sqrt{2}+\frac{2 x^{2}(2-\sqrt{2})}{4}$

$2 x^{2}=2+\sqrt{2}+x^{2}-\frac{x^{2} \sqrt{2}}{2}$

$x^{2}=2+\sqrt{2}-\frac{x^{2} \sqrt{2}}{2} / \cdot 2$

$2 x^{2}=4+2 \sqrt{2}-x^{2} \sqrt{2}$
$2 x^{2}+x^{2} \sqrt{2}=4+2 \sqrt{2}$

$x^{2}(2+\sqrt{2})=2(2+\sqrt{2})$

$x^{2}=2$

$x=\sqrt{2}$.

1 POINT

Note 2: After calculating $|A B|$, the student can obtain $x$ by applying the similarity of triangles.

Triangles $\triangle A D C$ and $\triangle A B E$ are similar by the $\mathrm{K}-\mathrm{K}$ similarity theorem (since $|\angle C D A|=|\angle A E B|=90^{\circ}$ and $\left.|\angle D A C|=|\angle E B A|=90^{\circ}-22.5^{\circ}=67.5^{\circ}\right)$.

1 POINT

Then,

$$
\frac{|A B|}{2}: v=y: x \text {, }
$$

or

$$
x \cdot \frac{\sqrt{2}}{2} \cdot \sqrt{2-\sqrt{2}}: \sqrt{2+\sqrt{2}}=x \cdot(\sqrt{2}-1): x
$$

1 POINT

It follows that:

$x \cdot \frac{\sqrt{2}}{2} \cdot \sqrt{2-\sqrt{2}}=\sqrt{2+\sqrt{2}} \cdot(\sqrt{2}-1) \quad / \cdot \sqrt{2-\sqrt{2}}$

1 POINT

$x \cdot \frac{\sqrt{2}}{2} \cdot(2-\sqrt{2})=\sqrt{(2+\sqrt{2})(2-\sqrt{2})} \cdot(\sqrt{2}-1)$

$x \cdot \frac{\sqrt{2}}{2} \cdot(2-\sqrt{2})=\sqrt{4-2} \cdot(\sqrt{2}-1)$

$x \cdot \frac{\sqrt{2}}{2} \cdot(2-\sqrt{2})=\sqrt{2} \cdot(\sqrt{2}-1)$

$x \cdot \frac{\sqrt{2}}{2} \cdot(2-\sqrt{2})=2-\sqrt{2}$

$x \cdot \frac{\sqrt{2}}{2}=1$

$x=\sqrt{2}$.

1 POINT