# Difference between revisions of "2016 AMC 8 Problems/Problem 25"

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==Solution 4: Inscribed Circle== | ==Solution 4: Inscribed Circle== | ||

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<asy> pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label("$B$",B,SW); label("$D$",D,SE); label("$A$",A,N); label("$M$",M,S); label("$C$",C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));</asy> | <asy> pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label("$B$",B,SW); label("$D$",D,SE); label("$A$",A,N); label("$M$",M,S); label("$C$",C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));</asy> |

## Revision as of 22:08, 16 February 2017

A semicircle is inscribed in an isosceles triangle with base and height so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?

## Contents

## Solution 1

Draw the altitude from the top of the triangle to its base, dividing the isosceles triangle into two right triangles with height and base . The Pythagorean triple -- tells us that these triangles have hypotenuses of .

Now draw an altitude of one of the smaller right triangles, starting from the foot of the first altitude we drew (which is also the center of the circle that contains the semicircle) and going to the hypotenuse of the right triangle. This segment is both an altitude of the right triangle as well as the radius of the semicircle (this is because tangent lines to circles, such as the hypotenuse touching the semicircle, are always perpendicular to the radii of the circles drawn to the point of tangency). Let this segment's length be .

The area of the entire isosceles triangle is , so the area of each of the two congruent right triangles it gets split into is . We can also find the area of one of the two congruent right triangles by using its hypotenuse as its base and the radius of the semicircle, the altitude we drew, as its height. Then the area of the triangle is . Thus we can write the equation , so , so .

## Solution 2 (Quicker Solution, basically solution one)

First, we draw a line perpendicular to the base of the triangle and cut it in half. The base of the resulting right triangle would be 8, and the height would be 15. Using the Pythagorean theorem, we can find the length of the hypotenuse, which would be 17. Using the two legs of the right angle, we can find the area of the right triangle, . times results in the radius, which is the height of the right triangle when using the hypotenuse as the base. The answer is .

## Solution 3: Similar Triangles

Let's call the triangle where and Let's say that is the midpoint of and is the point where is tangent to the semicircle. We could also use instead of because of symmetry.

Notice that and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by similarity, with and This similarity means that we can create a proportion: We plug in and After we multiply both sides by we get

(By the way, we could also use )

## Solution 4: Inscribed Circle

Solution by HydroQuantum

We'll call this triangle . Let the midpoint of base be . Divide the triangle in half by drawing a line from to . Half the base of is . The height is , which is given in the question. Using the Pythagorean Triple --, the length of each of the legs ( and ) is 17.

Reflect the triangle over its base. This will create an inscribed circle in a rhombus . Because , . Therefore .

The semiperimeter of the rhombus is . Since the area of is , the area of the rhombus is twice that, which is .

The Formula for the Incircle of a Quadrilateral is = . Substituting the semiperimeter and area into the equation, . Solving this, = .