2019 AMC 12B Problems/Problem 12
Contents
Problem
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 
?
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Solution 1
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
Solution 2
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
$= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}
, and$ (Error compiling LaTeX. Unknown error_msg)\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$.
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 11 |
Followed by Problem 13 |
| 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 12 Problems and Solutions | |
