Difference between revisions of "1987 AIME Problems/Problem 6"
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== Problem == | == Problem == | ||
| − | Rectangle <math>\displaystyle ABCD</math> is divided into four parts of equal area by five segments as shown in the figure, where <math>\displaystyle XY = YB + BC + CZ = ZW = WD + DA + AX</math>, and <math>\displaystyle PQ</math> is parallel to <math>\displaystyle AB</math>. Find the length of <math>\displaystyle AB</math> (in cm) if <math>\displaystyle BC = 19</math> cm and <math>\displaystyle PQ = 87</math> cm. | + | [[Rectangle]] <math>\displaystyle ABCD</math> is divided into four parts of equal [[area]] by five [[line segment | segments]] as shown in the figure, where <math>\displaystyle XY = YB + BC + CZ = ZW = WD + DA + AX</math>, and <math>\displaystyle PQ</math> is [[parallel]] to <math>\displaystyle AB</math>. Find the [[length]] of <math>\displaystyle AB</math> (in cm) if <math>\displaystyle BC = 19</math> cm and <math>\displaystyle PQ = 87</math> cm. |
[[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 | + | Since <math>XY = WZ</math> and <math>PQ = PQ</math> and the [[area]] of the [[trapezoid]]s <math>\displaystyle PQZW</math> and <math>\displaystyle PQYX</math> are the same, the heights of the trapezoids are the same. Thus both are equal to <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 /> |
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>. | 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>. | ||
Revision as of 18:26, 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. Thus both are equal to 
. 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 | ||
