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 <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, 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 07:46, 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 |
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