2014 AMC 10A Problems/Problem 22
Contents
[hide]- 1 Problem
- 2 Solution 1 (Trigonometry)
- 3 Solution 2 (No Trigonometry)
- 4 Solution 3 Quick Construction (No Trigonometry)
- 5 Solution 4 (No Trigonometry)
- 6 Solution 5
- 7 Solution 6 (Pure Euclidian Geometry)
- 8 Solution 7 (Pure Euclidian Geometry)
- 9 Solution 8 (Trigonometry)
- 10 Video Solution by Richard Rusczyk
- 11 See Also
Problem
In rectangle ,
and
. Let
be a point on
such that
. What is
?
Solution 1 (Trigonometry)
Note that . (It is important to memorize the sin, cos, and tan values of
and
.) Therefore, we have
. Since
is a
triangle,
Solution 2 (No Trigonometry)
Let be a point on line
such that points
and
are distinct and that
. By the angle bisector theorem,
. Since
is a
right triangle,
and
. Additionally,
Now, substituting in the obtained values, we get
and
. Substituting the first equation into the second yields
, so
. Because
is a
triangle,
.
~edited by ripkobe_745
Solution 3 Quick Construction (No Trigonometry)
Reflect over line segment
. Let the point
be the point where the right angle is of our newly reflected triangle. By subtracting
to find
, we see that
is a
right triangle. By using complementary angles once more, we can see that
is a
angle, and we've found that
is a
right triangle. From here, we can use the
properties of a
right triangle to see that
Solution 4 (No Trigonometry)
Let be a point on
such that
. Then
Since
,
is isosceles.
Let . Since
is
, we have
Since is isosceles, we have
. Since
, we have
Thus
and
.
Finally, by the Pythagorean Theorem, we have
~ Solution by Nafer
~ Edited by TheBeast5520
Note from williamgolly: When you find DE, note how ADE is congruent to a 30-60-90 triangle and you can easily find AE from there
Solution 5
First, divide all side lengths by to make things easier. We’ll multiply our answer by
at the end.
Call side length
. Using the Pythagorean Theorem, we can get side
is
.
The double angle identity for sine states that: So,
We know
. In triangle
,
and
. Substituting these in, we get our equation:
which simplifies to
Now, using the quadratic formula to solve for .
Because the length
must be close to one, the value of
will be
.
We can now find
=
and use it to find
.
.
To find
, we can use the Pythagorean Theorem with sides
and
, OR we can notice that, based on the two side lengths we know,
is a
triangle. So
.
Finally, we must multiply our answer by ,
.
.
~AWCHEN01
Solution 6 (Pure Euclidian Geometry)
We are going to use pure Euclidian geometry to prove .
Reflect rectangle along line
. Let the square be
as shown. Construct equilateral triangle
.
Because ,
, and
,
by
.
So, ,
.
Because ,
,
,
.
by
.
So, . By the reflection,
.
This solution is inspired by AoPS "Introduction to Geometry" page 226 problem 8.22, and page 433 problem 16.42.
Solution 7 (Pure Euclidian Geometry)
We are going to use pure Euclidian geometry to prove .
Construct equilateral triangle , and let
be the height of
.
,
,
,
.
by
.
,
,
, by
.
So, .
,
,
,
,
.
by
.
So,
Solution 8 (Trigonometry)
All trigonometric functions in this solution are in degrees. We know so
Let
, then
. By the definition of sine,
Squaring both sides,
Cross-multiplying,
Simplifying,
Let
. Notice that
so
.
is then
Recall that
which we now know is
Therefore
Rationalizing the denominator,
Which by difference of squares reduces to
so
.
is then
and since we know
, by the Pythagorean theorem,
. The answer is
An alternate way to finish: since we know the lengths of and
, we can figure out that
and therefore
. Hence
is isosceles and
.
~JH. L
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=-GBvCLSfTuo
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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