2017 AMC 8 Problems/Problem 22
In the right triangle , , , and angle is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?
We can reflect triangle over line This forms the triangle and a circle out of the semicircle. We can see that our circle is the incircle of We can use a formula for finding the radius of the incircle. The area of a triangle . The area of is The semiperimeter is Simplifying Our answer is therefore
Asymptote diagram by Mathandski
Let the center of the semicircle be . Let the point of tangency between line and the semicircle be . Angle is common to triangles and . By tangent properties, angle must be degrees. Since both triangles and are right and share an angle, is similar to . The hypotenuse of is , where is the radius of the circle. (See for yourself) The short leg of is . Because ~ , we have and solving gives
Let the tangency point on be . Note By Power of a Point, Solving for gives
Let us label the center of the semicircle and the point where the circle is tangent to the triangle . The area of = the areas of + , which means . So, it gives us .
Solution 5 (Pythagorean Theorem)
We can draw another radius from the center to the point of tangency. This angle, , is . Label the center , the point of tangency , and the radius .
Since is a kite, then . Also, . By the Pythagorean Theorem, . Solving, .
Solution 6 (Basic Trigonometry)
If we reflect triangle over line , we will get isosceles triangle . By the Pythagorean Theorem, we are capable of finding out that the . Hence, . Therefore, as of triangle , the radius of its inscribed circle
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