Difference between revisions of "2014 AMC 10B Problems/Problem 15"
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===Solution 2=== | ===Solution 2=== | ||
− | WLOG, let <math>AD = 1</math> and <math>BC = 2</math>. Furthermore, drop an an altitude from <math>F</math> to <math>CD</math>, which meets <math>CD</math> at <math>X</math>. Since <math>\angle ADC</math> is right and has been trisected, it follows that <math>\triangle ADE</math> and <math>\triangle DXF</math> are both <math>30-60-90</math> triangles. Therefore, <math>AE = \frac{\sqrt{3}}{3}</math>, and <math>DX = AF = \sqrt{3}</math>. Hence, it follows that <math>EF = \sqrt{3} - \frac{\sqrt{3}}{3}= \frac{2\sqrt{3}}{3}</math>. Since <math>\angle ADE</math> is right, the height and base of <math>\triangle DEF</math> are <math>1</math> and <math> \frac{2\sqrt{3}}{3}</math>, respectively. Thus, the area of <math>\triangle DEF</math> is <math>\frac{\sqrt{3}}{3}</math>, and the area of rectengle <math>ABCD</math> is <math>2</math>, so the ratio beween the area of <math>\triangle DEF</math> and <math>ABCD</math> is <math>\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}</math>. | + | WLOG, let <math>AD = 1</math> and <math>BC = 2</math>. Furthermore, drop an an altitude from <math>F</math> to <math>CD</math>, which meets <math>CD</math> at <math>X</math>. Since <math>\angle ADC</math> is right and has been trisected, it follows that <math>\triangle ADE</math> and <math>\triangle DXF</math> are both <math>30-60-90</math> triangles. Therefore, <math>AE = \frac{\sqrt{3}}{3}</math>, and <math>DX = AF = \sqrt{3}</math>. Hence, it follows that <math>EF = \sqrt{3} - \frac{\sqrt{3}}{3}= \frac{2\sqrt{3}}{3}</math>. Since <math>\angle ADE</math> is right, the height and base of <math>\triangle DEF</math> are <math>1</math> and <math> \frac{2\sqrt{3}}{3}</math>, respectively. Thus, the area of <math>\triangle DEF</math> is <math>\frac{\sqrt{3}}{3}</math>, and the area of rectengle <math>ABCD</math> is <math>2</math>, so the ratio beween the area of <math>\triangle DEF</math> and <math>ABCD</math> is <math>\boxed{\textbf{(A) }\frac{\sqrt{3}}{6}}</math>. Note that we are able to assume that <math>AD=1</math> and <math>BC = 2</math> because we were asked to find the ratio between two areas. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}} | {{AMC10 box|year=2014|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:06, 1 January 2018
Problem
In rectangle , and points and lie on so that and trisect as shown. What is the ratio of the area of to the area of rectangle ?
Solution
Solution 1
Let the length of be , so that the length of is and .
Because is a rectangle, , and so . Thus is a right triangle; this implies that , so . Now drop the altitude from of , forming two triangles.
Because the length of is , from the properties of a triangle the length of is and the length of is thus . Thus the altitude of is , and its base is , so its area is .
To finish, .
Solution 2
WLOG, let and . Furthermore, drop an an altitude from to , which meets at . Since is right and has been trisected, it follows that and are both triangles. Therefore, , and . Hence, it follows that . Since is right, the height and base of are and , respectively. Thus, the area of is , and the area of rectengle is , so the ratio beween the area of and is . Note that we are able to assume that and because we were asked to find the ratio between two areas.
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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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