2017 AIME II Problems/Problem 10
Problem
Rectangle 
has side lengths 
and 
. Point 
is the midpoint of 
, point 
is the trisection point of 
closer to 
, and point 
is the intersection of 
and 
. Point 
lies on the quadrilateral 
, and 
bisects the area of . Find the area of 
.
Solution
![[asy] pair A,B,C,D,M,n,O,P; A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); fill(C--D--P--cycle,lightgray); draw(A--B--C--D--cycle); draw(C--M); draw(D--n); draw(B--P); draw(D--P); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,W); label("$N$",n,N); label("$O$",O,(-0.5,1)); label("$P$",P,N); dot(A); dot(B); dot(C); dot(D); dot(M); dot(n); dot(O); dot(P); label("28",(0,42)--(28,42),N); label("56",(28,42)--(84,42),N); [/asy]](http://latex.artofproblemsolving.com/b/1/5/b15dc0e88c799e645a58cb1b0f7e03cf1abade7c.png)
Impose a coordinate system on the diagram where point 
is the origin. Therefore 
, 
, 
, and 
. Because is a midpoint and is a trisection point, 
and 
. The equation for line 
is 
and the equation for line 
is 
, so their intersection, point , is 
. Using the shoelace formula on quadrilateral , or or drawing diagonal 
and using 
, we find that its area is 
. Therefore the area of triangle 
is 
. Using 
, we get 
. Simplifying, we get 
. This means that the x-coordinate of 
. Since P lies on , you can solve and get that the y-coordinate of is 
. Therefore the area of 
is 
.
Solution Altered By conantwiz2023 and IronicNinja~
See Also
| 2017 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
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