Difference between revisions of "1959 AHSME Problems/Problem 36"
(created solution page) |
(diagram) |
||
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | import geometry; | ||
+ | |||
+ | size(10cm); | ||
+ | |||
+ | point A = (0,0); | ||
+ | point B = (17/20,17*sqrt(3)/20); | ||
+ | point C = (8,0); | ||
+ | triangle ABC = triangle(A,B,C); | ||
+ | |||
+ | // Triangle ABC | ||
+ | draw(ABC); | ||
+ | dot(A); | ||
+ | label("A",A,SW); | ||
+ | dot(B); | ||
+ | label("B",B,NW); | ||
+ | dot(C); | ||
+ | label("C",C,SE); | ||
+ | |||
+ | // Labels | ||
+ | markscalefactor = 0.1; | ||
+ | draw(anglemark(C,A,B)); | ||
+ | label("$60^{\circ}$", A, (1.5,1.5)); | ||
+ | label("$80$", midpoint(A--C), S); | ||
+ | label("$x$", midpoint(A--B), NW); | ||
+ | label("$90-x$", midpoint(B--C), NE); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | |||
<math>\boxed{\textbf{(D) 17}}</math> | <math>\boxed{\textbf{(D) 17}}</math> | ||
− | |||
== See also == | == See also == | ||
{{AHSME 50p box|year=1959|num-b=35|num-a=37}} | {{AHSME 50p box|year=1959|num-b=35|num-a=37}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:04, 21 July 2024
Problem
The base of a triangle is , and one side of the base angle is . The sum of the lengths of the other two sides is . The shortest side is:
Solution
See also
1959 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 35 |
Followed by Problem 37 | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.