Difference between revisions of "2016 AMC 8 Problems/Problem 23"
(Added another solution) |
m (small change) |
||
Line 9: | Line 9: | ||
Now, <math>\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) }\ 120}</math>. | Now, <math>\angle{CED}=m\angle{AEC}+m\angle{AEB}+m\angle{BED} = 30^{\circ}+60^{\circ}+30^{\circ} = 120^{\circ}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) }\ 120}</math>. | ||
− | ==Solution 2== | + | ==Solution 2 (Trig)== |
Let <math>r</math> be the radius of both circles (we are given that they are congruent). Let's drop the altitude from <math>E</math> onto segment <math>AB</math> and call the intersection point <math>F</math>. Notice that <math>F</math> is the midpoint of <math>A</math> and <math>B</math>, which means that <math>AF = BF = \frac{r}{2}</math>. Also notice that <math>\triangle{EAB}</math> is equilateral, which means we can use the Pythagorean Theorem to get <math>EF = \frac{r\sqrt{3}}{2}</math>. | Let <math>r</math> be the radius of both circles (we are given that they are congruent). Let's drop the altitude from <math>E</math> onto segment <math>AB</math> and call the intersection point <math>F</math>. Notice that <math>F</math> is the midpoint of <math>A</math> and <math>B</math>, which means that <math>AF = BF = \frac{r}{2}</math>. Also notice that <math>\triangle{EAB}</math> is equilateral, which means we can use the Pythagorean Theorem to get <math>EF = \frac{r\sqrt{3}}{2}</math>. | ||
− | Now let's apply trigonometry. Let <math>\theta = \angle{CEF}</math>. We can see that <math>\tan\theta = \frac{CF}{EF} = \frac{\frac{3r}{2}}{\frac{r\sqrt{3}}{2}} = \sqrt{3}</math>. This means <math>m\angle{CEF} = \frac{\pi}{3}</math>. However, this is not the answer. The question is asking for <math>m\angle{CED}</math>. Notice that <math>\angle{CEF}\cong\angle{DEF}</math>, which means <math>m\angle{CED} = 2m\angle{CEF}</math>. Thus, <math>\angle{CED} = 2\cdot\frac{\pi}{3} = \frac{2\pi}{3} = 120^{\circ}</math>. | + | Now let's apply trigonometry. Let <math>\theta = \angle{CEF}</math>. We can see that <math>\tan\theta = \frac{CF}{EF} = \frac{\frac{3r}{2}}{\frac{r\sqrt{3}}{2}} = \sqrt{3}</math>. This means <math>m\angle{CEF} = \frac{\pi}{3}</math>. However, this is not the answer. The question is asking for <math>m\angle{CED}</math>. Notice that <math>\angle{CEF}\cong\angle{DEF}</math>, which means <math>m\angle{CED} = 2m\angle{CEF}</math>. Thus, <math>\angle{CED} = 2\cdot\frac{\pi}{3} = \frac{2\pi}{3} = 120^{\circ}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) }\ 120}</math>. |
==Video Solution== | ==Video Solution== |
Revision as of 22:52, 21 August 2020
Problem 23
Two congruent circles centered at points and each pass through the other circle's center. The line containing both and is extended to intersect the circles at points and . The circles intersect at two points, one of which is . What is the degree measure of ?
Solution 1
Observe that is equilateral. Therefore, . Since is a straight line, we conclude that . Since (both are radii of the same circle), is isosceles, meaning that . Similarly, .
Now, . Therefore, the answer is .
Solution 2 (Trig)
Let be the radius of both circles (we are given that they are congruent). Let's drop the altitude from onto segment and call the intersection point . Notice that is the midpoint of and , which means that . Also notice that is equilateral, which means we can use the Pythagorean Theorem to get .
Now let's apply trigonometry. Let . We can see that . This means . However, this is not the answer. The question is asking for . Notice that , which means . Thus, . Therefore, the answer is .
Video Solution
https://youtu.be/WJ0Hodj0h2o - Happytwin
See Also
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.