2025 AIME II Problems/Problem 14

Solution 1(Coordinates and Bashy Algebra

By drawing our the triangle, I set A to be (0, 0) in the coordinate plane. I set C to be (x, 0) and B to be (0, y). I set K to be (a, b) and L to be (c, d). Then, since all of these distances are 14, I used coordinate geometry to set up the following equations: $a^{2}$ + $b^{2}$ = 196; $a^{2}$ + $(b - y)^{2}$ = 196; $(a - c)^{2}$ + $(b - d)^{2}$ = 196; $c^{2}$ + $d^{2}$ = 196; $(c - x)^{2}$ + $d^{2}$ . = 196. Notice by merging the first two equations, the only possible way for it to work is if $b - y$ = $-b$ which means $y = 2b$ . Next, since the triangle is right, and we know one leg is $2b$ as $y = 2b$ , the other leg, x, is $\sqrt{38^{2} - (2b)^{2}}$ .Then, plugging these in, we get a system of equations with 4 variables and 4 equations and solving, we get a = 2, b = 8 $\sqrt{3}$ , c = 13, d = 3 $\sqrt{3}$ . Now plugging in all the points and using the Pythagorean Theorem, we get the coordinates of the quadrilateral. By Shoelace, our area is 104 $\sqrt{3}$ . Thus, the answer is $\boxed{104}$ .

~ilikemath247365

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2025 AIME II Problems/Problem 14

Solution 1(Coordinates and Bashy Algebra