Crescents and triangles
Can you find a relationship between the area of the crescents and the area of the triangle?
Age
14 to 16
Challenge level
Triangle ABC is right angled at A and semi circles are drawn on all three sides producing two 'crescents'.
Can you show that the sum of the areas of the two crescents equals the area of triangle ABC?
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You do not even have to calculate areas to verify this result.
The trick with this question was not to make life difficult by working values out. As a consequence some of you managed much more elegant solutions than others. Here is an example based on that offered by Shu Cao (very well done). Jeongmin Lee, Rebecca Bartram (The Mount School, York) and Andrei Lazanu (School 205, Bucharest) should also be congratulated.
Using Pythagoras` theorem in
$ABC$,
$$\begin{eqnarray}a &=& \sqrt{b^2 + c^2} \\ \mbox{Area
of semicircle on BC} &=& \pi \frac {a^2}{4}\\ &=&
S1\\ \mbox{Area of semicircle on AC} &=& \pi \frac
{b^2}{4}\\ &=& S2 \\ \mbox{Area of semicircle on AB}
&=& \pi \frac{c^2}{4}\\ &=& S3\\ \mbox{Area of
crescents} &=& S2 + S3 + \mbox{Area ABC} - S1\\ &=&
\pi \frac {b^2}{4} +\pi \frac {c^2}{4} + \mbox{Area ABC} - \pi
\frac {a^2}{4}\\ &=& \pi \frac {b^2}{4} + \pi \frac{c^2}{4}
- \pi \frac {a^2}{4} + \mbox{Area ABC}\\ &=& \frac{\pi}{4}
\times ( b^2 + c^2 - a^2 ) + \mbox{Area ABC}\\ &=& 0 +
\mbox{Area ABC} \\ &=& \mbox{Area ABC}
\end{eqnarray}$$
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