Difference between revisions of "2014 AMC 10A Problems/Problem 22"
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Reflect <math>\triangle{ECB}</math> over line segment <math>\overline{CD}</math>. Let the point <math>F</math> be the point where the right angle is of our newly reflected triangle. By subtracting <math>90 - (15+15) = 60</math> to find <math>\angle ABF</math>, we see that <math>\triangle{ABC}</math> is a <math>30-60-90</math> right triangle. By using complementary angles once more, we can see that <math>\angle{EAD}</math> is a <math>60^\circ</math> angle, and we've found that <math>\triangle{EAD}</math> is a <math>30-60-90</math> right triangle. From here, we can use the <math>1-2-\sqrt{3}</math> properties of a <math>30-60-90</math> right triangle to see that <math>\overline{AE}=\boxed{\textbf{(E)}~20}.</math> | Reflect <math>\triangle{ECB}</math> over line segment <math>\overline{CD}</math>. Let the point <math>F</math> be the point where the right angle is of our newly reflected triangle. By subtracting <math>90 - (15+15) = 60</math> to find <math>\angle ABF</math>, we see that <math>\triangle{ABC}</math> is a <math>30-60-90</math> right triangle. By using complementary angles once more, we can see that <math>\angle{EAD}</math> is a <math>60^\circ</math> angle, and we've found that <math>\triangle{EAD}</math> is a <math>30-60-90</math> right triangle. From here, we can use the <math>1-2-\sqrt{3}</math> properties of a <math>30-60-90</math> right triangle to see that <math>\overline{AE}=\boxed{\textbf{(E)}~20}.</math> | ||
− | == Solution | + | == Solution 4 (No Trigonometry) == |
Let <math>F</math> be a point on <math>BC</math> such that <math>\angle{FEC}=60^{\circ}</math>. Then <cmath>\angle{BEF}=\angle{BEC}-\angle{FEC}=15^{\circ}</cmath> Since <math>\angle{BEF}=\angle{EBF}</math>, <math>\bigtriangleup{BFE}</math> is isosceles. | Let <math>F</math> be a point on <math>BC</math> such that <math>\angle{FEC}=60^{\circ}</math>. Then <cmath>\angle{BEF}=\angle{BEC}-\angle{FEC}=15^{\circ}</cmath> Since <math>\angle{BEF}=\angle{EBF}</math>, <math>\bigtriangleup{BFE}</math> is isosceles. |
Revision as of 22:10, 30 December 2019
Contents
[hide]Problem
In rectangle ,
and
. Let
be a point on
such that
. What is
?
Solution 1 (Trigonometry)
Note that . (If you do not know the tangent half-angle formula, it is
). 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,
.
We see that
is a
triangle, leaving
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
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 |
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