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Solution to puzzle 143: Semicircle in a triangle

Skip restatement of puzzle.In triangleABC, side AB = 20, AC = 11, and BC = 13.  Find the diameter of the semicircle inscribed in ABC, whose diameter lies on AB, and that is tangent to AC and BC.


A number of approaches to this problem are possible.

Firstly, we draw a line from C to the center of the circle O.
Then draw radii OD and OE to the points of contact of tangents CB and CA, respectively.  The radii meet the tangents at right angles.

Triangle ABC, with inscribed semicircle.

Using Angle Bisector Theorem

By symmetry, CD = CE.
Clearly OD = OE = radius of semicircle.
Hence triangleODC is similar to triangleOEC, and so angleOCE = angleOCD.

Applying the Angle Bisector Theorem to triangleABC, we have AC/BC = AO/BO.
Hence 11/13 = AO/(20 − AO), and so 13AO = 11(20 − AO) = 220 − 11AO.
Therefore AO = 55/6.

Applying the law of cosines (also known as the cosine rule) to triangleABC, we have
132 = 112 + 202 − 2·11·20·cos A.
Hence cos A = 4/5, and so, by Pythagoras' Theorem, sin A = 3/5.

In triangleEAO, sin A = OE/AO.
Hence OE = (3/5) · (55/6) = 11/2.

Therefore, the diameter of the semicircle is 11 units.

Using Area of Triangles

Applying Heron's Formula to triangleABC,
area of ABC = square root[22 · (22 − 11) · (22 − 13) · (22 − 20)] = 66.

Let the radius of the semicircle be r, so that OD = OE = r.
Then area of triangleAOC = ½ · AC · r = 11r/2.
Similarly, area of triangleBOC = ½ · BC · r = 13r/2.

Since area of triangleABC = area of triangleAOC + area of triangleBOC, we have 66 = 12r, and hence r = 11/2.

Therefore, the diameter of the semicircle is 11 units.

Source: Original

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