Difference between revisions of "2019 AMC 12B Problems/Problem 12"
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Plugging these into <math>\dfrac{2-CD^2}{AD^2}</math>, we'll get | Plugging these into <math>\dfrac{2-CD^2}{AD^2}</math>, we'll get | ||
− | <math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(2-AD)^2}{AD^2} = \dfrac{-2+4AD-AD^2}{AD^2}</math> and | + | <math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(2-AD)^2}{AD^2} = \dfrac{-2+4AD-AD^2}{AD^2}</math> |
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<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(AD^2-2)}{AD^2} = \dfrac{4-AD^2}{AD^2}</math>. | <math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(AD^2-2)}{AD^2} = \dfrac{4-AD^2}{AD^2}</math>. |
Revision as of 23:48, 14 February 2019
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 ?
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
and
. .
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
and
.
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 |