Difference between revisions of "2014 AMC 10A Problems/Problem 22"
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Note that <math>\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. (If you do not know the tangent half-angle formula, it is <math>\tan \frac{\theta}2= \frac{1-\cos \theta}{\sin \theta}</math>). Therefore, we have <math>DE=10\sqrt 3</math>. Since <math>ADE</math> is a <math>30-60-90</math> triangle, <math>AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}</math> | Note that <math>\tan 15^\circ=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3</math>. (If you do not know the tangent half-angle formula, it is <math>\tan \frac{\theta}2= \frac{1-\cos \theta}{\sin \theta}</math>). Therefore, we have <math>DE=10\sqrt 3</math>. Since <math>ADE</math> is a <math>30-60-90</math> triangle, <math>AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}</math> | ||
− | ==Solution 2 ( | + | ==Solution 2 (No Trigonometry)== |
Let <math>F</math> be a point on line <math>\overline{CD}</math> such that points <math>C</math> and <math>F</math> are distinct and that <math>\angle EBF = 15^\circ</math>. By the angle bisector theorem, <math>\frac{\overline{BC}}{\overline{BF}} = \frac{\overline{CE}}{\overline{EF}}</math>. Since <math>\triangle BFC</math> is a <math>30-60-90</math> right triangle, <math>\overline{CF} = \frac{10\sqrt{3}}{3}</math> and <math>\overline{BF} = \frac{20\sqrt{3}}{3}</math>. Additionally, <cmath>\overline{CE} + \overline{EF} = \overline{CF} = \frac{10\sqrt{3}}{3}</cmath>Now, substituting in the obtained values, we get <math>\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{\overline{CE}}{\overline{EF}} \Rightarrow \frac{2\sqrt{3}}{3}\overline{CE} = \overline{EF}</math> and <math>\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}</math>. Substituting the first equation into the second yields <math>\frac{2\sqrt{3}}{3}\overline{CE} + \overline{CE} = \frac{10\sqrt{3}}{3} \Rightarrow \overline{CE} = 20 - 10\sqrt{3}</math>, so <math>\overline{DE} = 10\sqrt{3}</math>. Because <math>\triangle ADE</math> is a <math>30-60-90</math> triangle, <math>\overline{AE} = \boxed{\textbf{(E)}~20}</math>. | Let <math>F</math> be a point on line <math>\overline{CD}</math> such that points <math>C</math> and <math>F</math> are distinct and that <math>\angle EBF = 15^\circ</math>. By the angle bisector theorem, <math>\frac{\overline{BC}}{\overline{BF}} = \frac{\overline{CE}}{\overline{EF}}</math>. Since <math>\triangle BFC</math> is a <math>30-60-90</math> right triangle, <math>\overline{CF} = \frac{10\sqrt{3}}{3}</math> and <math>\overline{BF} = \frac{20\sqrt{3}}{3}</math>. Additionally, <cmath>\overline{CE} + \overline{EF} = \overline{CF} = \frac{10\sqrt{3}}{3}</cmath>Now, substituting in the obtained values, we get <math>\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{\overline{CE}}{\overline{EF}} \Rightarrow \frac{2\sqrt{3}}{3}\overline{CE} = \overline{EF}</math> and <math>\overline{CE} + \overline{EF} = \frac{10\sqrt{3}}{3}</math>. Substituting the first equation into the second yields <math>\frac{2\sqrt{3}}{3}\overline{CE} + \overline{CE} = \frac{10\sqrt{3}}{3} \Rightarrow \overline{CE} = 20 - 10\sqrt{3}</math>, so <math>\overline{DE} = 10\sqrt{3}</math>. Because <math>\triangle ADE</math> is a <math>30-60-90</math> triangle, <math>\overline{AE} = \boxed{\textbf{(E)}~20}</math>. |
Revision as of 21:54, 16 August 2019
Contents
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 (Measuring)
If we draw rectangle and whip out a protractor, we can draw a perfect
, almost perfectly
off of
. Then we can draw
, and use a ruler to measure it.
We can clearly see that the
is
.
NOTE: this method is a last resort, and is pretty risky. Answer choice is also very close to
, meaning that we wouldn't be 100% sure of our answer. However, If we measure the angles of
, we can clearly see that it is a
triangle, which verifies our answer of
.
Solution 5 (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
~ Nafer
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.