Difference between revisions of "2019 AMC 12B Problems/Problem 12"
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==Solution 1.5 (Little bit of coordinate bash)== | ==Solution 1.5 (Little bit of coordinate bash)== | ||
− | After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math> | + | After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of our triangle by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively. |
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math> | By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math> |
Revision as of 19:25, 14 February 2019
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 our triangle 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
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 |