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

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Denote the bottom right vertex of the isosceles triangle to be <math>B</math> | Denote the bottom right vertex of the isosceles triangle to be <math>B</math> | ||

− | Denote the top | + | Denote the top vertex of the isosceles triangle to be <math>C</math> |

Drop an altitude from <math>C</math> to side <math>AB</math>. Denote the foot of intersection to be <math>D</math>. | Drop an altitude from <math>C</math> to side <math>AB</math>. Denote the foot of intersection to be <math>D</math>. | ||

− | By the Pythagorean Theorem, <math>AC=17</math> | + | By the Pythagorean Theorem, <math>AC=17</math>. |

− | Now, we see that <math>\sin{A}=\frac{15}{17}</math> | + | Now, we see that <math>\sin{A}=\frac{15}{17}</math>. |

− | This implies that <math>\sin{A}=\frac{r}{8}</math> (r=radius of semicircle) | + | This implies that <math>\sin{A}=\frac{r}{8}</math> (r=radius of semicircle). |

− | Hence, <math>r=\boxed{\frac{120}{17}}</math> | + | Hence, <math>r=\boxed{\frac{120}{17}}</math>. |

## Revision as of 11:52, 10 November 2018

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

First, we draw a line perpendicular to the base of the triangle and cut the triangle 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 2: 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

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, = .

## Solution 4: Inscribed Circle

Noting that we have a 8-15-17 triangle, we can find and Let , Then by similar triangles (or "Altitude on Hypotenuse") we have Thus, Now again by "Altitude on Hypotenuse”, Therefore

## Solution 5: Simple Trigonometry(10 second solve)

Denote the bottom left vertex of the isosceles triangle to be

Denote the bottom right vertex of the isosceles triangle to be

Denote the top vertex of the isosceles triangle to be

Drop an altitude from to side . Denote the foot of intersection to be .

By the Pythagorean Theorem, .

Now, we see that .

This implies that (r=radius of semicircle).

Hence, .