Difference between revisions of "2019 AMC 12B Problems/Problem 12"
(→Solution 2) |
(→Solution 2) |
||
Line 69: | Line 69: | ||
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math> | <math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math> | ||
<math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>. | <math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>. | ||
+ | |||
+ | So <math>2\sin(x+y)\cos(x+y) = [\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math>. | ||
==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 23:33, 14 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
.
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 .
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 |