2024 AMC 10A Problems/Problem 22

Problem

Let be the kite formed by joining two right triangles with legs and along a common hypotenuse. Eight copies of are used to form the polygon shown below. What is the area of triangle ?

Solution 1

Let be quadrilateral MNOP. Drawing line MO splits the triangle into . Drawing the altitude from N to point Q on line MO, we know NQ is , MQ is , and QO is .

Due to the many similarities present, we can find that AB is , and the height of is

AB is and the height of is .

Solving for the area of gives which is

~9897 (latex beginner here)

Solution 2

Let's start by looking at kite . We can quickly deduce based off of the side lengths that the kite can be split into -- triangles. Going back to the triangle ABC, focus on side AB. There are kites, they are all either reflected over the line AB or a line perpendicular to AB, meaning the length of AB can be split up into 4 equal parts.

Pick out the bottom-left kite, and we can observe that the kite and the triangle formed by the intersection of the kite and share a degree angle. (this was deduced from the -- triangles in the kite) The line AB and the right side of the kite are perpendicular, forming a degree angle. Because that is also a -- triangle with a hypotenuse of , so we find the length of AB to be , which is .

Then, we can drop an altitude from C to AB. We know that will be equivalent to the sum of the longer side of the kite and the shorter side of the triangle formed by the intersection of the kite and . (Look at the line formed on the left of C that drops down to AB if you are confused) We already have those values from the -- triangles, so we can just plug it into the triangle area formula, . We get ==> ==> ==>

~YTH (Need help with Latex and formatting)

~WIP (Header)

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.