2021 Fall AMC 10B Problems/Problem 15
Contents
[hide]Problem
In square , points and lie on and , respectively. Segments and intersect at right angles at , with and . What is the area of the square?
Solution 1
Note that Then, it follows that Thus, Define to be the length of side then Because is the altitude of the triangle, we can use the property that Substituting the given lengths, we have Solving, gives and We eliminate the possibilty of because Thus, the side lengnth of the square, by Pythagorean Theorem, is Thus, the area of the sqaure is Thus, the answer is
~NH14
Solution 2
Solution in Progress ~KingRavi
Solution 3
We have that Thus, . Now, let the side length of the square be Then, by the Pythagorean theorem, Plugging all of this information in, we get Simplifying gives Squaring both sides gives We now set and get the equation From here, notice we want to solve for , as it is precisely or the area of the square. So we use the Quadratic formula, and though it may seem bashy, we hope for a nice cancellation of terms. It seems scary, but factoring from the square root gives us giving us the solutions We instantly see that is way too small to be an area of this square ( isn't even an answer choice, so you can skip this step if out of time) because then the side length would be and then, even the largest line you can draw inside the square (the diagonal) is which is less than (line ) And thus, must be , and our answer is
~wamofan
Solution 4 (Point-line distance formula)
Denote . Now tilt your head to the right and view and as the origin, -axis and -axis, respectively. In particular, we have points . Note that side length of the square is . Also equation of line is Because the distance from to line is also the side length , we can apply the point-line distance formula to get which reduces to . Since is positive, the last equations factors as . Now judging from the figure, we learn that . So . Therefore, the area of the square is . Choose .
~VensL.
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.