Difference between revisions of "1987 AIME Problems/Problem 6"
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===Solution 3=== | ===Solution 3=== | ||
− | Since <math>XY = YB + BC + CZ = ZW = WD + DA + AX</math>. Let <math>a = AX + DW = BY + CZ</math>. Since 2AB - 2a = XY = WZ, then <math>XY = AB - a</math>.Let <math>S</math> be the midpoint of <math>DA</math>, and <math>T</math> be the midpoint of <math>CB</math>. Since the area of <math>PQWZ</math> and <math>PQYX</math> are the same, then their heights are the same, and so <math>PQ</math> is [[equidistant]] from <math>AB</math> and <math>CD</math>. This means that <math>PS</math> is perpendicular to <math>DA</math>, and <math>QT</math> is perpendicular to <math>BC</math>. Therefore, <math>PSCW</math>, <math>PSAX</math>, <math>QZCT</math>, and <math>QYTB</math> are all trapezoids, and <math>QT = (AB - 87)/</math>2. This implies that <cmath>((a + 2((AB - 87)/2)/2) \cdot 19 = (((AB - a) + 87)/2) \cdot 19</cmath> <cmath>(a + AB - 87) = AB - a + 87</cmath> <cmath>2a = 174</cmath> <cmath>a = 87</cmath> Since <math>a + CB = XY</math>, <math>XY = 19 + 87 = 106, and AB = 106 + 87 = \boxed{193}</math>. | + | Since <math>XY = YB + BC + CZ = ZW = WD + DA + AX</math>. Let <math>a = AX + DW = BY + CZ</math>. Since <math>2AB - 2a = XY = WZ</math>, then <math>XY = AB - a</math>.Let <math>S</math> be the midpoint of <math>DA</math>, and <math>T</math> be the midpoint of <math>CB</math>. Since the area of <math>PQWZ</math> and <math>PQYX</math> are the same, then their heights are the same, and so <math>PQ</math> is [[equidistant]] from <math>AB</math> and <math>CD</math>. This means that <math>PS</math> is perpendicular to <math>DA</math>, and <math>QT</math> is perpendicular to <math>BC</math>. Therefore, <math>PSCW</math>, <math>PSAX</math>, <math>QZCT</math>, and <math>QYTB</math> are all trapezoids, and <math>QT = (AB - 87)/</math>2. This implies that <cmath>((a + 2((AB - 87)/2)/2) \cdot 19 = (((AB - a) + 87)/2) \cdot 19</cmath> <cmath>(a + AB - 87) = AB - a + 87</cmath> <cmath>2a = 174</cmath> <cmath>a = 87</cmath> Since <math>a + CB = XY</math>, <math>XY = 19 + 87 = 106</math>, and <math>AB = 106 + 87 = \boxed{193}</math>. |
== See also == | == See also == |
Latest revision as of 06:46, 1 June 2018
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
Solution 1
Since , and the areas of the trapezoids and are the same, then the heights of the trapezoids are the same. Thus both trapezoids have area . This number is also equal to one quarter the area of the entire rectangle, which is , so we have .
In addition, we see that the perimeter of the rectangle is , so .
Solving these two equations gives .
Solution 2
Let , , , and . First we drop a perpendicular from to a point on so . Since and and the areas of the trapezoids and are the same, the heights of the trapezoids are both .From here, we have that . We are told that this area is equal to . Setting these equal to each other and solving gives . In the same way, we find that the perpendicular from to is . So
Solution 3
Since . Let . Since , then .Let be the midpoint of , and be the midpoint of . Since the area of and are the same, then their heights are the same, and so is equidistant from and . This means that is perpendicular to , and is perpendicular to . Therefore, , , , and are all trapezoids, and 2. This implies that Since , , and .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.