# Problem of the Week Problem E and Solution Outside the Path of Totality

## Problem

Yannick used a camera with a solar filter to capture a solar eclipse. From their location, the Moon blocked some, but not all of the Sun, so they saw a partial solar eclipse. When Yannick enlarged and printed the photo, they noticed that the distance represented by segment $$XY$$ in the photo shown was $$70$$ cm, and that the diameters of the Sun and the Moon were both $$74$$ cm.

Determine the percentage of the Sun that is blocked by the Moon in Yannick’s photo, rounded to $$1$$ decimal place.

## Solution

Let the centres of the Sun and Moon in the photo be $$S$$ and $$M$$, respectively. We draw line segments $$SX$$, $$SY$$, $$MX$$, and $$MY$$. We draw line segment $$SM$$ which intersects $$XY$$ at $$T$$. The shaded region represents the portion of the Sun that is blocked by the Moon.

In our diagram we have drawn $$S$$ and $$M$$ inside the shaded region, but we need to prove they actually lie inside this region before we can proceed with the area calculation. Since each of the circles has radius $$74 \div 2 = 37$$, then $$SX=SY=MX=MY=37$$. It follows that $$\triangle SXY$$ is isosceles. Similarly, $$\triangle SXM$$ and $$\triangle SYM$$ are congruent. Thus, $$\angle XSM = \angle YSM$$. Since $$\triangle SXY$$ is isosceles and $$SM$$ bisects $$\angle XSY$$, then $$SM$$ is perpendicular to $$XY$$ at $$T$$, and $$XT=TY=70 \div 2 = 35$$. By the Pythagorean Theorem in $$\triangle STX$$, $ST = \sqrt{SX^2-XT^2} = \sqrt{37^2-35^2} = 12$ Thus, $$SM=2 \times ST = 2 \times 12 = 24$$. Since $$SM$$ is smaller than the radius of the circles, it follows that $$S$$ and $$M$$ must lie inside the shaded region.

Now we can proceed with the area calculation. By symmetry, the area of the shaded region on each side of $$XY$$ will be the same. The area of the shaded region on the right side of $$XY$$ equals the area of acute sector $$XSY$$ of the left circle minus the area of $$\triangle SXY$$. These areas are striped in the following diagrams.

First we find the area of $$\triangle SXY$$, which is $$\frac{1}{2}(XY)(ST)=\frac{1}{2}(70)(12)=420$$.

Next we find the area of sector $$XSY$$. Using the cosine law in $$\triangle SXY$$, \begin{aligned} XY^2 &= SX^2 + SY^2 - 2(SX)(SY)\cos (\angle XSY)\\ 70^2 &= 37^2 + 37^2 - 2(37)(37)\cos (\angle XSY)\\ 2162 &= -2738\cos (\angle XSY)\\ \angle XSY &= \cos^{-1}\left(-\frac{2162}{2738}\right)\approx 142.15\degree \end{aligned} Thus, the area of sector $$XSY$$ is equal to $$\frac{142.15\degree}{360\degree}\pi (37)^2$$.

Then the area of the shaded region is equal to $$2 \left(\frac{142.15\degree}{360\degree}\pi (37)^2 - 420 \right)$$.

Finally, we calculate this area as a percentage of the area of the entire circle to obtain the following: $\frac{2 \left(\frac{142.15\degree}{360\degree}\pi (37)^2 - 420 \right)}{\pi (37)^2} \approx 59.4 \%$

It follows that the percentage of the Sun that is blocked by the Moon in Yannick’s photo is equal to approximately $$59.4 \%$$.