Difference between revisions of "2019 AMC 12B Problems/Problem 12"
Sevenoptimus (talk | contribs) (Thoroughly cleaned up the first two solutions for LaTeX, formatting, grammar, and clarity, as well as adding a lot more detail; also removed the third solution which was just the same as the first solution) |
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Thus we have the lengths <math>DE=1+\frac{1}{2\sqrt{2}}</math> (it is just the <math>y</math>-coordinate) and <math>AE = 1-\frac{1}{2\sqrt{2}}</math>. By simple trigonometry in <math>\triangle DAE</math>, we now find <cmath>\sin{\left(\angle BAD\right)} = \frac{\left(1+\frac{1}{2\sqrt{2}}\right)}{\left(\frac{3}{2}\right)} = \frac{\left(2+\frac{1}{\sqrt{2}}\right)}{3} = \frac{2\sqrt{2}+1}{3\sqrt{2}}</cmath> and <cmath>\cos{\left(\angle BAD\right)} = \frac{\left(1-\frac{1}{2\sqrt{2}}\right)}{\left(\frac{3}{2}\right)} = \frac{\left(2-\frac{1}{\sqrt{2}}\right)}{3} = \frac{2\sqrt{2}-1}{3\sqrt{2}}</cmath> just as before. We can then use the double angle formula (as in Solution 1) to deduce <math>\sin{\left(2\angle BAD\right)} = \boxed{\textbf{(D) } \frac{7}{9}}</math>. | Thus we have the lengths <math>DE=1+\frac{1}{2\sqrt{2}}</math> (it is just the <math>y</math>-coordinate) and <math>AE = 1-\frac{1}{2\sqrt{2}}</math>. By simple trigonometry in <math>\triangle DAE</math>, we now find <cmath>\sin{\left(\angle BAD\right)} = \frac{\left(1+\frac{1}{2\sqrt{2}}\right)}{\left(\frac{3}{2}\right)} = \frac{\left(2+\frac{1}{\sqrt{2}}\right)}{3} = \frac{2\sqrt{2}+1}{3\sqrt{2}}</cmath> and <cmath>\cos{\left(\angle BAD\right)} = \frac{\left(1-\frac{1}{2\sqrt{2}}\right)}{\left(\frac{3}{2}\right)} = \frac{\left(2-\frac{1}{\sqrt{2}}\right)}{3} = \frac{2\sqrt{2}-1}{3\sqrt{2}}</cmath> just as before. We can then use the double angle formula (as in Solution 1) to deduce <math>\sin{\left(2\angle BAD\right)} = \boxed{\textbf{(D) } \frac{7}{9}}</math>. | ||
+ | ==Solution 3 (easier finish to Solution 1)== | ||
+ | Again, use Pythagorean Theorem to find that <math>AD=\frac{3}{2}</math> and <math>CD=\frac{1}{2}</math>. Let <math>\angle DAC=\theta</math>. Note that we want <cmath>\sin{90+2\theta}=\cos{2\theta}</cmath> | ||
+ | which is easy to compute: <math></math>\cos{\theta}=\frac{2\sqrt{2}}{3}\implies \cos{2\theta}=2(\frac{8}{9})-1=\boxed{\textbf{(D) }\frac{7}{9} | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}} | {{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}} |
Revision as of 20:38, 7 March 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
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: $$ (Error compiling LaTeX. ! Missing $ inserted.)\cos{\theta}=\frac{2\sqrt{2}}{3}\implies \cos{2\theta}=2(\frac{8}{9})-1=\boxed{\textbf{(D) }\frac{7}{9}
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 |