Difference between revisions of "1987 AIME Problems/Problem 6"
m (→Solution 2) |
m (→Solution 2) |
||
Line 14: | Line 14: | ||
===Solution 2=== | ===Solution 2=== | ||
− | Let <math>YB=a</math>, <math>CZ=b</math>, <math>AX=c</math>, and <math>WD=d</math>. First we drop a perpendicular from <math>Q</math> to a point <math>R</math> on <math>BC</math> so <math>QR=h</math>. Since <math>XY = WZ</math> and <math>PQ = PQ</math> and the [[area]]s of the [[trapezoid]]s <math>PQZW</math> and <math>PQYX</math> are the same, the heights of the trapezoids are both <math>\frac{19}{2}</math>.From here, we have that <math>[BYQZC]=\frac{a+h}{2}*19/2+\frac{b+h}{2}*19/2=19/2* \frac{a+b+2h}{2}</math>. We are told that this area is equal to <math>[PXYQ]=\frac{19}{2}* \frac{XY+87}{2}=\frac{19}{2}* \frac{a+b+106}{2}</math>. Setting these equal to each other and solving gives <math> | + | Let <math>YB=a</math>, <math>CZ=b</math>, <math>AX=c</math>, and <math>WD=d</math>. First we drop a perpendicular from <math>Q</math> to a point <math>R</math> on <math>BC</math> so <math>QR=h</math>. Since <math>XY = WZ</math> and <math>PQ = PQ</math> and the [[area]]s of the [[trapezoid]]s <math>PQZW</math> and <math>PQYX</math> are the same, the heights of the trapezoids are both <math>\frac{19}{2}</math>.From here, we have that <math>[BYQZC]=\frac{a+h}{2}*19/2+\frac{b+h}{2}*19/2=19/2* \frac{a+b+2h}{2}</math>. We are told that this area is equal to <math>[PXYQ]=\frac{19}{2}* \frac{XY+87}{2}=\frac{19}{2}* \frac{a+b+106}{2}</math>. Setting these equal to each other and solving gives <math>h=53</math>. In the same way, we find that the perpendicular from <math>P</math> to <math>AD</math> is <math>53</math>. So <math>AB=53*2+87=\boxed{193}</math> |
===Solution 3=== | ===Solution 3=== |
Revision as of 21:54, 25 November 2016
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 2AB - 2a = XY = WZ, 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 , .
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.