Y by ostriches88, gracemoon124, clarkculus, golue3120, Rounak_iitr
In the figure below 
is a rectangle with 
and 
. Point 
lies 
, point 
lies on 
, and 
is a right angle. The areas of 
and 
are equal. What is the area of 
?
![[asy]
pair X = (0, 0);
pair W = (0, 4);
pair Y = (8, 0);
pair Z = (8, 4);
label("$X$", X, dir(180));
label("$W$", W, dir(180));
label("$Y$", Y, dir(0));
label("$Z$", Z, dir(0));
draw(W--X--Y--Z--cycle);
dot(X);
dot(Y);
dot(W);
dot(Z);
pair M = (2, 0);
pair A = (8, 3);
label("$A$", A, dir(0));
dot(M);
dot(A);
draw(W--M--A--cycle);
markscalefactor = 0.05;
draw(rightanglemark(W, M, A));
label("$M$", M, dir(-90));
[/asy]](//latex.artofproblemsolving.com/6/6/8/668d1b9909d619b769900471cd1fd94a582d030b.png)
is a rectangle with 
and 
. Point 
lies 
, point 
lies on 
, and 
is a right angle. The areas of 
and 
are equal. What is the area of 
?![[asy]
pair X = (0, 0);
pair W = (0, 4);
pair Y = (8, 0);
pair Z = (8, 4);
label("$X$", X, dir(180));
label("$W$", W, dir(180));
label("$Y$", Y, dir(0));
label("$Z$", Z, dir(0));
draw(W--X--Y--Z--cycle);
dot(X);
dot(Y);
dot(W);
dot(Z);
pair M = (2, 0);
pair A = (8, 3);
label("$A$", A, dir(0));
dot(M);
dot(A);
draw(W--M--A--cycle);
markscalefactor = 0.05;
draw(rightanglemark(W, M, A));
label("$M$", M, dir(-90));
[/asy]](http://latex.artofproblemsolving.com/6/6/8/668d1b9909d619b769900471cd1fd94a582d030b.png)
This post has been edited 13 times. Last edited by peace09, Nov 13, 2024, 6:56 PM
,
(and 
)
,
,
,![$[AEF]=\tfrac{2\sqrt{5}\cdot3\sqrt{5}}{2}=\boxed{\textbf{(C) }15}$](http://latex.artofproblemsolving.com/4/b/8/4b81e12b8bf9d1e36762858c13ea4d5eaaf493a6.png)
.
are 
triangles. 
and 
,
Then, 
Since 
by 
similarity, we get 
so 
Now, 
so 
Thus, we have that 
and 
for an answer of 
and 
Thus, 
and 
This gives us ![$[WXM]=2x$](http://latex.artofproblemsolving.com/8/1/1/811dcbea963aa58243cd7ff5c299a9c892bf2dbe.png)
and ![$[WZA]=4y.$](http://latex.artofproblemsolving.com/1/9/e/19e8c5fbf0de0494cea878a05ffb663222543c08.png)
We are given that ![$[WXM]=[WZA],$](http://latex.artofproblemsolving.com/7/8/c/78ce3c981bfaebf23b66b9f3d1f9fb7f3ada0338.png)
so 
Thus, ![$[WXM]=4y.$](http://latex.artofproblemsolving.com/f/3/5/f3593db3cec4f8cbb28967d631edc9b6887e492e.png)
![$[MYA]=\frac{(8-x)(4-y)}{2},$](http://latex.artofproblemsolving.com/9/6/b/96b72947f4067d376cefc6ef79ca8e4b409fdc6a.png)
or 
That expression can be simplified to 
![$[WXM]=[WZA]=4y,$](http://latex.artofproblemsolving.com/b/1/6/b16f182abf3b74c3e08de025b8d08ed29b79e9e3.png)
so the total area inside the rectangle but outside ![$[WMA]$](http://latex.artofproblemsolving.com/a/1/1/a11400bdda6f35bc7d6b9e1f096f7592d23a637f.png)
is 
must be a positive integer (because of the answer choices), so the only way that 
can equal something that is 32 minus one of the answer choices is if 
So, the area inside the rectangle but outside 
