Inscribed Triangle

Prove that the total area of the regions 1 and 2 below equals the area of the triangle. Regions 3 and 4 are parts of the semi-circle in which the triangle is inscribed, and \overline{AB} and \overline{BC} are diameters for the other two semi-circles.

Source: Julia Robinson Mathematics Festival

PROOF
Area 1 = area of semi-circle AEB- area 3

=\frac{\pi}{2}\left (\frac{AB}{2}\right )^2- area 3

=\frac{\pi}{2}\times \frac{AB^2}{4}- area 3        (1)

Area 2 = area of semi-circle BFC- area 4

=\frac{\pi}{2}\left (\frac{BC}{2}\right )^2- area 4

=\frac{\pi}{2}\times \frac{BC^2}{4}- area 4        (2)

Adding Eq. (1) and Eq. (2) yields

Area 1 + Area 2 = \left (\frac{\pi}{2}\times \frac{AB^2}{4}\right )+\left (\frac{\pi}{2}\times \frac{BC^2}{4}\right )- area 3 - area 4

=\frac{\pi}{2}\left (\frac{AB^2+BC^2}{4}\right )- area 3 - area 4

Triangle ABC is a right triangle because inscribed angle B intercepts an arc of 180^\circ.

By the Pythagorean theorem, AB^2+BC^2=AC^2. Hence, it follows that

Area 1 + Area 2 = \frac{\pi}{2}\left (\frac{AC^2}{4}\right )- area 3 - area 4

The right hand-side of the above equation is precisely the area of triangle ABC.
DONE

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , , , . Bookmark the permalink.

One Response to Inscribed Triangle

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