Difference between revisions of "2014 AMC 10A Problems/Problem 22"
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Let <math>CF=x</math>. Since <math>\bigtriangleup{FEC}</math> is <math>60^{\circ}-90^{\circ}-30^{\circ}</math>, we have <math>EF=\frac{2}{\sqrt{3}}x</math> | Let <math>CF=x</math>. Since <math>\bigtriangleup{FEC}</math> is <math>60^{\circ}-90^{\circ}-30^{\circ}</math>, we have <math>EF=\frac{2}{\sqrt{3}}x</math> | ||
− | Since <math>\bigtriangleup{BFE}</math> is isosceles, we have <math>BF=EF=\frac{2}{\sqrt{3}}x</math>. Since <math>BF+FC=BF</math>, we have <cmath>\frac{2}{\sqrt{3}}x+x=10 \Longrightarrow x=20\sqrt{3}-30</cmath> Thus <math>EC=\frac{1}{\sqrt{3}} | + | Since <math>\bigtriangleup{BFE}</math> is isosceles, we have <math>BF=EF=\frac{2}{\sqrt{3}}x</math>. Since <math>BF+FC=BF</math>, we have <cmath>\frac{2}{\sqrt{3}}x+x=10 \Longrightarrow x=20\sqrt{3}-30</cmath> Thus <math>EC=\frac{1}{\sqrt{3}}BC=20-10\sqrt{3}</math> and <math>DE=DC-EC=20-EC=10\sqrt{3}</math>. |
Finally, by the Pythagorean Theorem, we have <cmath>AE=\sqrt{AD^2+DE^2}=\sqrt{10^2+(10\sqrt{3})^2}=20 \boxed{\mathrm{(E)}}</cmath> | Finally, by the Pythagorean Theorem, we have <cmath>AE=\sqrt{AD^2+DE^2}=\sqrt{10^2+(10\sqrt{3})^2}=20 \boxed{\mathrm{(E)}}</cmath> | ||
− | ~ Nafer | + | ~ Solution by Nafer |
+ | |||
+ | ~ Edited by TheBeast5520 | ||
==See Also== | ==See Also== |
Revision as of 19:23, 17 November 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 (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
~ Solution by Nafer
~ Edited by TheBeast5520
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.