Difference between revisions of "1991 AHSME Problems/Problem 19"
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== Solution 4 (Really simple similar triangles)== | == Solution 4 (Really simple similar triangles)== | ||
− | + | By Pythagoras on right triangle <math>ABD</math>, <math>BD = 13</math>. Draw the line through <math>D</math> parallel to <math>CE</math>. Let this line intersect <math>AC</math> at <math>F</math>. | |
+ | [asy] | ||
+ | unitsize(0.3 cm); | ||
+ | |||
+ | pair A, B, C, D, E, F; | ||
+ | |||
+ | A = (0,3); | ||
+ | B = (4,0); | ||
+ | C = (0,0); | ||
+ | D = scale(12/5)*rotate(90)*(B - A) + A; | ||
+ | E = (D + reflect(B,C)*(D))/2; | ||
+ | F = extension(A,C,D, D + C - E); | ||
+ | |||
+ | draw(A--C--E--D); | ||
+ | draw(A--B--D--cycle); | ||
+ | draw(A--F--D); | ||
+ | |||
+ | label("<math>A</math>", A, W); | ||
+ | label("<math>B</math>", B, S); | ||
+ | label("<math>C</math>", C, SW); | ||
+ | label("<math>D</math>", D, NE); | ||
+ | label("<math>E</math>", E, SE); | ||
+ | label("<math>F</math>", F, NW); | ||
+ | [/asy] | ||
+ | Since <math>\angle DAB = 90^\circ</math>, <math>\angle FAD + \angle CAB = 90^\circ</math>. Hence, right triangles <math>ABC</math> and <math>DAF</math> are similar. Then | ||
+ | <cmath>\frac{AF}{4} = \frac{12}{5},</cmath> | ||
+ | so <math>AF = 48/5</math>. | ||
+ | |||
+ | Since quadrilateral <math>CEDF</math> is a rectangle, <math>DE = CF = AC + AF = 3 + 48/5 = 63/5</math>. Therefore, | ||
+ | <cmath>\frac{DE}{DB} = \frac{63/5}{13} = \frac{63}{65}.</cmath> | ||
+ | Then <math>m = 63</math>, <math>n = 65</math>, and <math>m + n = 63 + 65 = \boxed{128}</math>. The answer is (B). | ||
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Revision as of 21:41, 23 November 2020
Contents
Problem
Triangle has a right angle at and . Triangle has a right angle at and . Points and are on opposite sides of . The line through parallel to meets extended at . If where and are relatively prime positive integers, then
Solution 1
Solution by e_power_pi_times_i
Let be the point such that and are parallel to and , respectively, and let and . Then, . So, . Simplifying , and . Therefore , and . Checking, is the answer, so . The answer is .
Solution 2
Solution by Arjun Vikram
Extend lines and to meet at a new point . Now, we see that . Using this relationship, we can see that , (so ), and the ratio of similarity between and is . This ratio gives us that . By the Pythagorean Theorem, . Thus, , and the answer is .
Solution 3 (Trig)
We have and Now we are trying to find Now we use the angle sum identity, which states Using this identity yields
Solution 4 (Really simple similar triangles)
By Pythagoras on right triangle , . Draw the line through parallel to . Let this line intersect at . [asy] unitsize(0.3 cm);
pair A, B, C, D, E, F;
A = (0,3); B = (4,0); C = (0,0); D = scale(12/5)*rotate(90)*(B - A) + A; E = (D + reflect(B,C)*(D))/2; F = extension(A,C,D, D + C - E);
draw(A--C--E--D); draw(A--B--D--cycle); draw(A--F--D);
label("", A, W); label("", B, S); label("", C, SW); label("", D, NE); label("", E, SE); label("", F, NW); [/asy] Since , . Hence, right triangles and are similar. Then so .
Since quadrilateral is a rectangle, . Therefore, Then , , and . The answer is (B).
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See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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.