Difference between revisions of "2019 AMC 12B Problems/Problem 12"
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− | Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>. | + | Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>. (Note: You could set up variables such as X, 2-X to denote the side lengths |
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>. | Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>. | ||
Revision as of 00:34, 15 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 . (Note: You could set up variables such as X, 2-X to denote the side lengths 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
.
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,
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 |