Difference between revisions of "2014 AMC 10A Problems/Problem 22"
(→Solution 10 Law of Cosines) |
(→Solution 10 Law of Cosines) |
||
Line 151: | Line 151: | ||
This equals, <math>(AE)^2=20^2+200(\sqrt{3}-1)^2-200(\sqrt{3}-1)^2)</math>. | This equals, <math>(AE)^2=20^2+200(\sqrt{3}-1)^2-200(\sqrt{3}-1)^2)</math>. | ||
− | The 200(\sqrt{3}-1)^2) cancel out each other. We are left with <math>(AE)^2=20^2</math>. So, <math>(AE)=20</math>. | + | The <math>200(\sqrt{3}-1)^2)</math> cancel out each other. We are left with <math>(AE)^2=20^2</math>. So, <math>(AE)=20=\boxed{\textbf{(E)} \: 20}</math>. |
== Video Solution by Richard Rusczyk == | == Video Solution by Richard Rusczyk == |
Revision as of 18:14, 29 October 2023
Contents
- 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 Solution 9
- 11 Solution 10 Law of Cosines
- 12 Video Solution by Richard Rusczyk
- 13 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,
Note: Similar to previous Solution
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
Solution 9
Similar to the others except that you find the base of the 15-75-90 triangle knowing that the side opposite side to 75 is 10, finding that you can subtract to find the base of the rectangle of the diagonal we are trying to find.
~YBSuburbanTea
Solution 10 Law of Cosines
The ratio between the side lengths of a 15, 75, 90 triangle are . Therefore,
. Using triangular ratios,
.
.
From the law of cosines, .
So, .
At this stage, it is easier if we do not expand the numbers yet. Further simplifying, we have .
Now, we have .
Combing like terms, .
This equals, .
The cancel out each other. We are left with
. So,
.
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 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.