Difference between revisions of "2019 AMC 12B Problems/Problem 12"
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==Solution 3 (easier finish to Solution 1)== | ==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> | 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: < | + | which is easy to compute: <cmath>\cos{\theta}=\frac{2\sqrt{2}}{3}\implies \cos{2\theta}=2(\frac{8}{9})-1=\boxed{\textbf{(D) }\frac{7}{9}</cmath> |
+ | |||
==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 21:38, 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:
\[\cos{\theta}=\frac{2\sqrt{2}}{3}\implies \cos{2\theta}=2(\frac{8}{9})-1=\boxed{\textbf{(D) }\frac{7}{9}\] (Error compiling LaTeX. Unknown error_msg)
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 |