Difference between revisions of "1959 AHSME Problems/Problem 36"
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== 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> | ||
+ | |||
+ | Let the triangle be <math>\triangle ABC</math> with <math>\measuredangle BAC=60^{\circ}</math> and <math>AC=80</math>, as in the diagram. Let <math>AB=x</math>. from the problem, we know that <math>BC=90-x</math>. Now, we can apply the [[Law of Cosines]] on <math>\triangle ABC</math> to solve for <math>x</math>: | ||
+ | \begin{align*} | ||
+ | (90-x)^2 &= 6400+x^2-160\cos(60^{\circ}) \\ | ||
+ | x^2-180x+8100 &= 6400+x^2-80x \\ | ||
+ | -100x &= -1700 \\ | ||
+ | x &= 17 | ||
+ | \end{align*} | ||
+ | Because the problem asks for the shortest side of the triangle and <math>90-x=73>17</math>, our answer is <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}} | ||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 16:11, 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
Let the triangle be with
and
, as in the diagram. Let
. from the problem, we know that
. Now, we can apply the Law of Cosines on
to solve for
:
\begin{align*}
(90-x)^2 &= 6400+x^2-160\cos(60^{\circ}) \\
x^2-180x+8100 &= 6400+x^2-80x \\
-100x &= -1700 \\
x &= 17
\end{align*}
Because the problem asks for the shortest side of the triangle and
, our answer is
.
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.