考题19 2015 AMC 10A

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校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AMC10数学竞赛真题以及官方解答吧：

Problem 19

The isosceles right triangle  has right angle at  and area . The rays trisecting  intersect  at  and . What is the area of ?

Solution 1

 can be split into a  right triangle and a  right triangle by dropping a perpendicular from  to side . Let  be where that perpendicular intersects .

Because the side lengths of a  right triangle are in ratio , .

Because the side lengths of a  right triangle are in ratio  and  + , .

Setting the two equations for  equal to each other, .

Solving gives .

The area of .

 is congruent to , so their areas are equal.

A triangle's area can be written as the sum of the figures that make it up, so .

.

Solving gives , so the answer is 

Solution 2

The area of  is , and so the leg length of   is  Thus, the altitude to hypotenuse , , has length  by  right triangles. Now, it is clear that , and so by the Exterior Angle Theorem,  is an isosceles  triangle. Thus,  by the Half-Angle formula, and so the area of  is . The answer is thus 

Solution 3

Because the area of triangle  is , and the triangle is right and isosceles, we can quickly see that the leg length of the triangle  is 5. If we put the triangle on the coordinate plane, with vertex  at the origin, and the hypotenuse in the first quadrant, we can use slope-intercept form and tangents to get three lines that intersect at the origin, , and . Then, you can use the distance formula to get the length of . The height is just , so the area is just 

Solution 4

Just like with Solution 1, we drop a perpendicular from  onto , splitting it into a -- triangle and a -- triangle. We find that .

Now, since  by ASA, . Since, , . By the sine area formula, 

以上就是小编对AMC10数学竞赛真题以及解析的介绍，希望对你有所帮助，更多学习资料请持续关注AMC数学竞赛网！