## Task 4 - 250824

Let ABC be an isosceles right triangle with $C$ as the vertex of the right angle. Over $BC$ as the diameter, draw the semicircle that intersects $AB$ at a point $D$ between $A$ and $B$ (see figure).

a) Prove that the line through $D$ and the midpoint $E$ of $BC$ is perpendicular to $BC$!

b) Calculate what percentage of the area of triangle $ABC$ is not covered by the semicircle! The desired percentage should be given to one decimal place.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0757.jpg?height=392&width=375&top_left_y=632&top_left_x=1383)

Hint: Use the approximation $\pi \approx 3.142$!
a) Since $BC$ is the diameter of the semicircle, the midpoint $E$ of $BC$ is also the midpoint of the semicircle. The points $B$ and $D$ lie on the semicircle; thus, $EB = ED$.

This implies $\angle DBE = \angle BDE$ (base angles in the isosceles triangle $BDE$).

On the other hand, $\angle DBE = \angle ABC = 45^\circ$ (base angles in the isosceles right triangle $ABC$). By the angle sum property, it follows that

$$
\angle BED = 180^\circ - (\angle DBE + \angle BDE) = 90^\circ
$$

which means the claim $DE \perp BC$.

b) Since $\angle BED = \angle BCA = 90^\circ$, by the converse of the corresponding angles theorem for intersected lines, $ED \parallel CA$, so the quadrilateral $ACED$ is a trapezoid.

The height of the trapezoid is the radius $r = EC$ of the semicircle; the lengths of the parallel sides of the trapezoid are $ED = r$ and $CA = BC = 2r$. The area $J$ of the part of the triangle $ABC$ not covered by the semicircle can be represented as the difference between the areas of the trapezoid $ACED$ and a quarter circle with radius $r$:

$$
J = \frac{r + 2r}{2} r - \frac{\pi}{4} r^2 = \frac{r^2}{6} (6 - \pi)
$$

The area of the triangle $ABC$ is

$$
F = \frac{1}{2} \cdot 2r \cdot 2r = 2r^2
$$

The desired percentage is therefore

$$
p = \frac{100J}{F} = \frac{25r^2 (6 - \pi)}{2r^2} = 12.5 (6 - \pi)
$$

From $\pi \approx 3.142$, it follows that $p \approx 35.7$ to one decimal place. Thus, $35.7\%$ of the area of the triangle $ABC$ is not covered by the semicircle.

Solutions of the 2nd Round 1985 adopted from [5]

### 5.27.3 3rd Round 1985, Class 8