Difference between revisions of "1988 AHSME Problems/Problem 16"
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==Solution== | ==Solution== | ||
+ | Let <math>\triangle ABC</math> have side length <math>s</math> and <math>\triangle A'B'C'</math> have side length <math>t</math>. Thus the altitude of <math>\triangle ABC</math> is <math>\frac{s\sqrt{3}}{2}</math>. Now observe that this altitude is made up of three parts: the distance from <math>BC</math> to <math>B'C'</math>, plus the altitude of <math>\triangle A'B'C'</math>, plus a top part which is equal to the length of the diagonal line from the bottom-left corner of <math>\triangle ABC</math> to the bottom left corner of <math>\triangle A'B'C'</math> (as an isosceles trapezium is formed with parallel sides <math>AB</math> and <math>A'B'</math>, and legs <math>AA'</math> and <math>BB'</math>). We drop a perpendicular from <math>B'</math> to <math>BC</math>, which meets <math>BC</math> at <math>D</math>. <math>\triangle BDB'</math> has angles <math>30^{\circ}</math>, <math>60^{\circ}</math>, and <math>90^{\circ}</math>, and the vertical side is that distance from <math>BC</math> to <math>B'C'</math>, which is given as <math>\frac{1}{6} \times \frac{s\sqrt{3}}{2} = \frac{s\sqrt{3}}{12}</math>, so that by the length relationships in a 30-60-90 triangle,, the length of the diagonal line is <math>\frac{s\sqrt{3}}{12} \times 2 = \frac{s\sqrt{3}}{6}.</math> Thus using the "altitude in three parts" idea, we get <math>\frac{s\sqrt{3}}{2} = \frac{s\sqrt{3}}{12} + \frac{t\sqrt{3}}{2} + \frac{s\sqrt{3}}{6} \implies \frac{s\sqrt{3}}{4} = \frac{t\sqrt{3}}{2} \implies t = \frac{1}{2}s.</math> Thus the sides of <math>\triangle A'B'C'</math> are half as long as <math>\triangle ABC</math>, so the area ratio is <math>(\frac{1}{2}) ^ {2} = \frac{1}{4}</math>, which is <math>\boxed{\text{C}}</math>. | ||
− | + | ~Johnxyz1(minorEdits) | |
== See also == | == See also == |
Latest revision as of 09:07, 7 September 2024
Problem
and are equilateral triangles with parallel sides and the same center, as in the figure. The distance between side and side is the altitude of . The ratio of the area of to the area of is
Solution
Let have side length and have side length . Thus the altitude of is . Now observe that this altitude is made up of three parts: the distance from to , plus the altitude of , plus a top part which is equal to the length of the diagonal line from the bottom-left corner of to the bottom left corner of (as an isosceles trapezium is formed with parallel sides and , and legs and ). We drop a perpendicular from to , which meets at . has angles , , and , and the vertical side is that distance from to , which is given as , so that by the length relationships in a 30-60-90 triangle,, the length of the diagonal line is Thus using the "altitude in three parts" idea, we get Thus the sides of are half as long as , so the area ratio is , which is .
~Johnxyz1(minorEdits)
See also
1988 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
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