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


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.

A circle with interior coloured grey covers a
large portion of another circle with interior coloured yellow. The two
circles have the same radius and their centres lie along the same
horizontal line. The circles intersect at points X and Y and a dashed
vertical line segment connects X and Y.

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


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.

Two overlapping circles with the same radius.
The left circle has centre S and the right circle has centre M and SM is
horizontal, and the circles intersect at X at the top and Y at the
bottom. The region enclosed by the right edge of the left circle and the
left edge of the right circle is shaded. This region contains segment XY
and points points S, T, and M.

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 \(59.4 \%\).