Difference between revisions of "1987 AHSME Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | <math>\triangle</math> DBE is similar to <math>\triangle</math> ABC by AA, so <math>\overline{DE}</math> = 1 by similarity, and <math>\overline{CE} = \overline{AD} = 2</math>, by subtraction. Thus the perimeter is <math>3+2+2+1 = 8</math>, or <math>\boxed{E}</math> | + | <math>\triangle</math> DBE is similar to <math>\triangle</math> ABC by AA, so <math>\overline{DE}</math> = 1 by similarity, and <math>\overline{CE} = \overline{AD} = 2</math>, by subtraction. Thus the perimeter is <math>3+2+2+1 = 8</math>, or <math>\boxed{E}</math>. -slackroadia |
== See also == | == See also == | ||
Revision as of 17:37, 23 April 2017
Problem
A triangular corner with side lengths 
is cut from equilateral triangle ABC of side length 
.
The perimeter of the remaining quadrilateral is
![[asy] draw((0,0)--(2,0)--(2.5,.87)--(1.5,2.6)--cycle, linewidth(1)); draw((2,0)--(3,0)--(2.5,.87)); label("3", (0.75,1.3), NW); label("1", (2.5, 0), S); label("1", (2.75,.44), NE); label("A", (1.5,2.6), N); label("B", (3,0), S); label("C", (0,0), W); label("D", (2.5,.87), NE); label("E", (2,0), S); [/asy]](http://latex.artofproblemsolving.com/7/6/a/76a745ea106ed38564b062962b8548574d1399c5.png)
Solution
DBE is similar to ABC by AA, so 
= 1 by similarity, and 
, by subtraction. Thus the perimeter is 
, or 
. -slackroadia
See also
| 1987 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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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