AMC数学竞赛8专为8年级及以下的初中学生设计，但近年来的数据显示，越来越多小学4-6年级的考生加入到AMC 8级别的考试行列中，而当这些学生能在成绩中取得“A”类标签，则是对孩子数学天赋的优势证明，不管是对美高申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC 8的官方真题以及官方解答吧：

Problem 25

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?

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

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 verted 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, 

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