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 . Solving gives . (Note: You could set up variables such as X, 2-X to denote the side lengths). Using the sine and cosine addition formulas on angles and (which requires finding their cosines as well),
Similarly, we can do this with the cosine addition formula to find to get . \Now, use on angle to get .
Note: This has been elaborated on a little, but it is still not fully complete. Feel free to elaborate further 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
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 .
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