2015 AMC 10A Problems/Problem 19
Contents
Problem
The isosceles right triangle has right angle at and area . The rays trisecting intersect at and . What is the area of ?
Solution 1 (No Trigonometry)
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
Note
Another way to get is that you assume to be equal to , as previously mentioned, and is equal to .
Solution 2 (Trigonometry)
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 (Analytical Geometry)
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 (Trigonometry)
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,
Solution 5 (Basic Trigonometry)
Prerequisite knowledge for this solution: the side ratios of a 30-60-90, and 45-45-90 right triangle.
We let point C be the origin. Since and trisect , this means and the equation of is (you can figure out that the tangent of 30 degrees gives ). Next, we can find A to be at and B at , so the equation of is . So we have the system:
By substituting values, we can arrive at , or . We multiply .
Dropping an altitude from E onto , and calling the intersection point G, we find that is a 45-45-90 triangle with a leg of . Thus, .
Dropping an altitude from C onto , and calling the intersection point H, we find that , and by the theorem of betweenness applied to H, E, and B, we get .
We are almost done. By symmetry, , so to find the area of the triangle CED, we only need to multiply HE by CH, . This is answer choice ~JH. L
Solution 6 (Law of Sines)
We know that the area of the right triangle is and that the two legs are equal, so we can easily tell that the length of the two legs is . Thus, the hypotenuse and Let's quickly define as the point that bisects and . Then, we can say that the area of the desired triangle is . Let Since is one of the trisecting points of Because Now, we can employ the Law of Sines. It tells us that . Plugging in our angle values, we get that It's easy to find that and that . Plugging in these values into our previous equation, we get Cross multiplying gets us and then we simplify like so: Now, using our definition of , we know that = . We want to put this under one common denominator, which is pretty simple to execute. That leaves us with Whew. That was longer than expected. Anyways, quick inspection tells us that so now we just have to do some simplifying to find the desired, . Let's do that now. (We need to take a quick conjugation break. Note that ) ~Nickelslordm
Video Solution:
https://www.youtube.com/watch?v=JWMIsCS0Ksk
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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