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 ?
Solution 1
Firstly, note by the Pythagorean Theorem in that . Now, the equal perimeter condition means that , since side is common to both triangles and thus can be discounted. This relationship, in combination with the Pythagorean Theorem in , gives . Hence , so , and thus .
Next, since , . Using the lengths found above, , and .
Thus, by the addition formulae for and , we have and
Hence, by the double angle formula for , .
Solution 2 (coordinate geometry)
We use the Pythagorean Theorem, as in Solution 1, to find and . Now notice that the angle between and the vertical (i.e. the -axis) is – to see this, drop a perpendicular from to which meets at , and use the fact that the angle sum of quadrilateral must be . Anyway, this implies that the line has slope , so since is the point and the length of is , has coordinates .
Thus we have the lengths (it is just the -coordinate) and . By simple trigonometry in , we now find and just as before. We can then use the double angle formula (as in Solution 1) to deduce .
Solution 3 (easier finish to Solution 1)
Again, use Pythagorean Theorem to find that and . Let . Note that we want
which is easy to compute:
\[\cos{\theta}=\frac{2\sqrt{2}}{3}\implies \cos{2\theta}=2(\frac{8}{9})-1=\boxed{\textbf{(D) }\frac{7}{9}\] (Error compiling LaTeX. ! File ended while scanning use of \boxed.)
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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All AMC 12 Problems and Solutions |