Difference between revisions of "2014 AMC 10A Problems/Problem 22"
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− | Because <math>GH=HE=HF</math>, | + | Because <math>GH=HE=HF</math>, <math>\angle GHF= \angle BEF=75^{\circ} + 75^{\circ} = 150^{\circ}</math>, <math>\angle GHE=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}</math>, <math>\angle GHE=\angle GHF</math>. |
<math>\triangle GHE \cong \triangle GHF</math> by <math>SAS</math> | <math>\triangle GHE \cong \triangle GHF</math> by <math>SAS</math> | ||
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This solution is inspired by AoPS "Introduction to Geometry" page 226 problem 8.22, and page 433 problem 16.42. | This solution is inspired by AoPS "Introduction to Geometry" page 226 problem 8.22, and page 433 problem 16.42. | ||
+ | |||
+ | |||
+ | ~isabelchen | ||
+ | |||
+ | ==Solution 7 (Pure Euclidian Geometry)== | ||
+ | |||
+ | [[File:Rectangle.PNG|700px]] | ||
+ | |||
+ | We are going to use pure Euclidian geometry to prove <math>AE=AB</math>. | ||
+ | |||
+ | Construct equilateral triangle <math>\triangle BEF</math>, and let <math>GF</math> be the height of <math>\triangle ABF</math>. | ||
+ | |||
+ | <math>\angle GBF=90^{\circ}-15^{\circ}-60^{\circ}=15^{\circ}</math>, <math>\angle GBF=\angle CBE</math>, <math>\angle BGF=\angle BCE=90^{\circ}</math>, <math>BF=BE</math>. | ||
+ | |||
+ | <math>\triangle BGF \cong \triangle BCE</math> by <math>AAS</math>. | ||
+ | |||
+ | |||
+ | <math>BG=BC=10, AG=20-10=10</math>, <math>AG=BG</math>, <math>GF=GF</math>, by <math>HL</math> <math>\triangle AGF \cong \triangle BGF</math>. | ||
+ | |||
+ | So, <math>AF=BF=EF</math> | ||
+ | |||
+ | |||
+ | <math>\angle AFB=75^{\circ}+75^{\circ}=150^{\circ}</math>, <math>\angle AFE=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}</math>, <math>\angle AFB=\angle AFE</math>, <math>AF=AF</math>, <math>BF=EF</math>. | ||
+ | |||
+ | <math>\triangle AFB \cong \triangle AFE</math> by <math>SAS</math> | ||
+ | |||
+ | |||
+ | So, <math>AE=AB=\boxed{\textbf{(E)}~20}</math> | ||
Revision as of 01:47, 2 October 2021
Contents
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.
~isabelchen
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,
~isabelchen
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.