# Solution to the Two Triangles and Three Concentric Semi-Circles Puzzle +-- {.image} [[TwoTrianglesandThreeConcentricSemiCircles.png:pic]] > Two right-angled triangles and three concentric semicircles. What fraction is yellow? =-- ## Solution by [[Similar Triangles]] +-- {.image} [[TwoTriThreeConSemiCirclesLabelled.png:pic]] =-- In the above diagram, point $O$ is the centre of the concentric semi-circles. Consider the smaller triangle $A D E$ and the two smaller circles. Point $F$ is where the side $A E$ touches the circle, so angle $O \hat{F} E$ is a right-angle as it is the [[angle between a radius and tangent]]. Triangles $O D E$, $O F E$, and $O FA$ are all [[right-angled triangles]] with $O D$ and $O F$ of the same length, and $O E$ and $O A$ of the same length. They are therefore [[congruent]]. They are also [[similar]] to triangle $A D E$ as triangles $O F A$ and $E D A$ are both [[right-angled triangles]] sharing the angle at $A$. The congruency shows that the length of $E A$ is twice that of $E D$, and so by similarity $O E$ has double the length of $O D$. Therefore the middle semi-circle has double the radius of the smaller one. Applying the same argument to the triangle $A B C$ shows that the outer semi-circle has double the radius of the middle semi-circle, and so four times that of the smaller semi-circle. Since [[area|area scales with the square]] of the length scale factor, the area of the outer semi-circle is $16$ times that of the smallest, while the middle semi-circle has area $4$ times that of the smallest. The white region therefore has area $4 - 1 = 3$ times that of the smallest and so the fraction that is shaded is $\frac{13}{16}$. ## Solution by [[Equilateral Triangles]] +-- {.image} [[TwoTriThreeConSemiCirclesReflected.png:pic]] =-- Reflecting the diagram to create whole circles shows that the original triangles are half of a triangle with the property that its [[circumcircle]] and [[in-circle]] are [[concentric]]. Such a triangle must be [[equilateral]], and the radius of its circumcircle is double that of its in-circle. The result then follows as above.