Difference between revisions of "2019 AMC 12B Problems/Problem 12"
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So <math>2\sin(x+y)\cos(x+y) = 2[\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math> | So <math>2\sin(x+y)\cos(x+y) = 2[\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math> | ||
<math>= \dfrac{2-CD^2}{AD^2}</math>. | <math>= \dfrac{2-CD^2}{AD^2}</math>. | ||
+ | Now we just need to figure out what the numerical answer is. | ||
+ | |||
+ | From the given information about the triangles' perimeters, we can deduce that <math>CD + AD = 2</math>. | ||
+ | Also, the Pythagorean theorem tell us that <math>CD^2 + 2 = AD^2</math>. | ||
+ | These two equations allow us to write <math>CD and CD^2</math> in terms of <math>AD</math> without redundancy: <math>CD = 2 - AD</math> and <math>CD^2 = AD^2 - 2</math>. | ||
+ | Plugging these into <math>\dfrac{2-CD^2}{AD^2}</math>, we'll get | ||
+ | <math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(2-AD)^2}{AD^2} = \dfrac{-2+4AD-AD^2}{AD^2}</math> and | ||
+ | <math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(AD^2-2)}{AD^2} = \dfrac{4-AD^2}{AD^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 23:47, 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
.
Now we just need to figure out what the numerical answer is.
From the given information about the triangles' perimeters, we can deduce that .
Also, the Pythagorean theorem tell us that
.
These two equations allow us to write
in terms of
without redundancy:
and
.
Plugging these into
, we'll get
and
.
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 |