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Geometry Difficulty 4.8 AIME Find the answer Canada

In the diagram, a figure is drawn on a $6 \times 8$ grid using eight semi-circles whose diameters are $AB$ , $BC$ , $CD$ , $DE$ , $EF$ , $FG$ , $GH$ , and $HA$ .Hide/Reveal Description of Diagram for Question 24 Eight semi-circles connect to form a closed shape on a $6 \times 8$ grid. With the bottom left corner of the grid having coordinates $(0,0)$ . the coordinates of the end points of the eight diameters are as follows: $A(1,4)$ and $B(3,5)$ $B(3,5)$ and $C(5,5)$ $C(5,5)$ and $D(7,4)$ $D(7,4)$ and $E(7,2)$ $E(7,2)$ and $F(5,2)$ $F(5,2)$ and $G(3,2)$ $G(3,2)$ and $H(1,2)$ $H(1,2)$ and $A(1,4)$ Suppose that the area of the figure is $x$ and that $y$ is the closest integer to $100x$ . What is the sum of the digits of
$y$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Draw line segments from $A$ to $B$ , $B$ to $C$ , $C$ to $D$ , $D$ to $E$ , $E$ to $F$ , $F$ to $G$ , $G$ to $H$ , and $H$ to $A$ , as shown. [[IMAGE0]] The line segments $BC$ , $DE$ , $EF$ , $FG$ , $GH$ , and $HA$ each have a length of $2$ units. Hence, the radii of the semicircles with these diameters are all $1$ , and the areas of the circles with these diameters are all $\dfrac{1}{2}\pi(1)^2=\dfrac{\pi}{2}$ . The line segment $AB$ is the hypotenuse of a triangle with legs of length $1$ and $2$ . By the Pythagorean Theorem, the length of $AB$ is $\sqrt{1^2+2^2}=\sqrt{5}$ . The radius of the semicircle with diameter $AB$ is $\dfrac{\sqrt{5}}{2}$ , so its area is $\dfrac{1}{2}\pi\left(\dfrac{\sqrt{5}}{2}\right)^2=\dfrac{5\pi}{8}$ . By similar reasoning, the area of the semicircle with diameter $CD$ is also $\dfrac{5\pi}{8}$ . The area of the figure can be computed as the area of hexagon $ABCDEH$ plus the areas of the semicircles with diameters $AB$ , $CD$ , $EF$ , and $GH$ , minus the areas of the semicircles with diameters $BC$ , $DE$ , $FG$ , and $AH$ . We have already computed the areas of the semicircles, so we now need to compute the area of hexagon $ABCDEH$ . This hexagon can be viewed as a $\$3\times$
6 rectangle with two “corners” removed. These “corners” are right-angled triangles with hypotenuses

Figure for this problem AB $and$ CD. The legs of these two triangles have length

Figure for this problem 1 $and$ 2, so their areas are each

Figure for this problem $\dfrac{1}{2}\times1\times2=1$ $. Thus, the area of hexagon$ ABCDEH $is$ $3\times 6-2\times$ 1=16. Using the areas of the semicircles computed earlier, we can now compute the area of the figure as

Figure for this problem $16 +\frac{5\pi}{8} + \frac{5\pi}{8} + \frac{\pi}{2} + \frac{\pi}{2} - \frac{\pi}{2} - \frac{\pi}{2} - \frac{\pi}{2} - \frac{\pi}{2}=16+\dfrac{\pi}{4}\approx 16.78539$ $Thus,$ $x\approx$ 16.78539 $, so$ $100x\approx$ 1678.539 $. Rounding to the nearest integer, we get$ y=1679 $, so the answer is$ 1+6+7+9=23$.

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