Difference between revisions of "1987 AIME Problems/Problem 6"
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[[Image:AIME_1987_Problem_6.png]] | [[Image:AIME_1987_Problem_6.png]] | ||
== Solution == | == Solution == | ||
+ | Since <math>XY = WZ</math> and <math>PQ = PQ</math> and the area of the trapezoids <math>\displaystyle PQZW</math> and <math>\displaystyle PQYX</math> are the same, the heights of the trapezoids are the same, or <math>\frac{19}{2}</math>. Extending <math>PQ</math> to <math>AD</math> and <math>BC</math> at <math>P'</math> and <math>Q'</math>, we split the [[rectangle]] into two congruent parts, such that <math>\displaystyle DWPP' + CZQQ' = PQZW</math> (this can be proved through a quick subtraction of areas).<br /><br /> | ||
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+ | Therefore, <math>\frac{1}{2} \cdot \frac{19}{2}(CZ + DW)(AB - PQ) = \frac{1}{2} \cdot \frac{19}{2}(PQ)(AB - (DW + CZ))</math>, which boils down to <math>CZ + DW = 87</math> (the same can reasoning can be repeated for the bottom half to yield <math>AX + BY = 87</math>). Notice that <math>AB = \frac{WZ + XY + (AX + BY)}{2} = \frac{(WD + DA + AX) + (YB + BC + CZ) + (CZ + DW) + (AX + BY)}{2} = \frac{87 \cdot 4 + 19 \cdot 2}{2} = 193</math>. | ||
== See also == | == See also == |
Revision as of 18:18, 15 February 2007
Problem
Rectangle is divided into four parts of equal area by five segments as shown in the figure, where
, and
is parallel to
. Find the length of
(in cm) if
cm and
cm.
Solution
Since and
and the area of the trapezoids
and
are the same, the heights of the trapezoids are the same, or
. Extending
to
and
at
and
, we split the rectangle into two congruent parts, such that
(this can be proved through a quick subtraction of areas).
Therefore, , which boils down to
(the same can reasoning can be repeated for the bottom half to yield
). Notice that
.
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |