2019 AMC 12B Problems/Problem 12
Right triangle with right angle at is constructed outwards on the hypotenuse of isosceles right triangle with leg length , as shown, so that the two triangles have equal perimeters. What is ?
Observe that the "equal perimeter" part implies that . A quick Pythagorean chase gives . Use the sine addition formula on angles and (which requires finding their cosines as well), and this gives the sine of . Now, use on angle to get .
Feel free to elaborate if necessary.
Solution 1.5 (Little bit of coordinate bash)
After using Pythagorean to find and , we can instead notice that the angle between the y-coordinate and is degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point , we can then proceed to find the height and base of this new triangle (defined by where is the intersection of the altitude and ) by coordinate-bashing, which turns out to be and respectively.
By double angle formula and difference of squares, it's easy to see that our answer is
~Solution by MagentaCobra
Let and , so .
By the double-angle formula, . To write this in terms of and , we can say that we are looking for .
Using trigonometric addition and subtraction formulas, we know that
So . Now we just need to figure out what the numerical answer is.
From the given information about the triangles' perimeters, we can deduce that . Also, the Pythagorean theorem tell us that . These two equations allow us to write in terms of without redundancy: and .
Plugging these into , we'll get
If we set these equal to each other, now there is an algebraic equation that can be easily solved:
Now that we know what is equal to, we can also figure out . Thus, .
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