Difference between revisions of "1990 AHSME Problems/Problem 20"
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== Solution == | == Solution == | ||
− | <math>\fbox{C}</math> | + | |
+ | Label the angles as shown in the diagram. Since <math>\angle DEC</math> forms a linear pair with <math>\angle DEA</math>, <math>\angle DEA</math> is a right angle. | ||
+ | |||
+ | <asy> | ||
+ | pair A = (0,0), B = (7,4.2), C = (10, 0), D = (3, -5), E = (3, 0), F = (7,0); | ||
+ | draw(A--B--C--D--cycle,dot); | ||
+ | draw(A--E--F--C,dot); | ||
+ | draw(D--E--F--B,dot); | ||
+ | |||
+ | markscalefactor = 0.075; | ||
+ | |||
+ | draw(rightanglemark(B, A, D)); | ||
+ | draw(rightanglemark(D, E, A)); | ||
+ | draw(rightanglemark(B, F, A)); | ||
+ | draw(rightanglemark(D, C, B)); | ||
+ | draw(rightanglemark(D, E, C)); | ||
+ | draw(rightanglemark(B, F, C)); | ||
+ | MP("A",(0,0),W); | ||
+ | MP("B",(7,4.2),N); | ||
+ | MP("C",(10,0),E); | ||
+ | MP("D",(3,-5),S); | ||
+ | MP("E",(3,0),N); | ||
+ | MP("F",(7,0),S); | ||
+ | </asy> | ||
+ | |||
+ | Let <math>\angle DAE = \angle \alpha</math> and <math>\angle BAF = \angle \beta</math>. | ||
+ | |||
+ | Since <math>\angle \alpha + \angle \beta = 90^\circ</math>, and <math>\angle \alpha + \angle BAF = 90^\circ</math>, then <math>\angle \beta = \angle BAF</math>. By the same logic, <math>\angle ABF = \angle \alpha</math>. | ||
+ | |||
+ | |||
+ | As a result, <math>\triangle AED \sim \triangle BFA</math>. By the same logic, <math>\triangle CFB \sim \triangle DEC</math>. | ||
+ | |||
+ | Then, <math>\frac{BF}{AF} = \frac{3}{5}</math>, and <math>\frac{CF}{BF} = \frac{5}{7}</math>. | ||
+ | |||
+ | Then, <math>7CF = 5BF</math>, and <math>5BF = 3AF</math>. | ||
+ | |||
+ | By the transitive property, <math>7CF = 3AF</math>. <math>AC = AF + CF = 10</math>, and plugging in, we get <math>CF = 3</math>. | ||
+ | |||
+ | Finally, plugging in to <math>\frac{CF}{BF} = \frac{5}{7}</math>, we get <math>BF = 4.2</math> <math>\fbox{C}</math> | ||
== See also == | == See also == |
Revision as of 12:55, 22 February 2018
Problem
In the figure is a quadrilateral with right angles at and . Points and are on , and and are perpendicual to . If and , then
Solution
Label the angles as shown in the diagram. Since forms a linear pair with , is a right angle.
Let and .
Since , and , then . By the same logic, .
As a result, . By the same logic, .
Then, , and .
Then, , and .
By the transitive property, . , and plugging in, we get .
Finally, plugging in to , we get
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.