2021 AMC 12B Problems/Problem 24
Problem
Let 
be a parallelogram with area 
. Points 
and 
are the projections of 
and 
respectively, onto the line 
and points 
and 
are the projections of 
and 
respectively, onto the line 
See the figure, which also shows the relative locations of these points.
![[asy] size(350); defaultpen(linewidth(0.8)+fontsize(11)); real theta = aTan(1.25/2); pair A = 2.5*dir(180+theta), B = (3.35,0), C = -A, D = -B, P = foot(A,B,D), Q = -P, R = foot(B,A,C), S = -R; draw(A--B--C--D--A^^B--D^^R--S^^rightanglemark(A,P,D,6)^^rightanglemark(C,Q,D,6)); draw(B--R^^C--Q^^A--P^^D--S,linetype("4 4")); dot("$A$",A,dir(270)); dot("$B$",B,E); dot("$C$",C,N); dot("$D$",D,W); dot("$P$",P,SE); dot("$Q$",Q,NE); dot("$R$",R,N); dot("$S$",S,dir(270)); [/asy]](http://latex.artofproblemsolving.com/d/3/5/d357d6211c4c672e68a32c940ef9ef86b8947dd2.png)
Suppose 
and 
and let 
denote the length of 
the longer diagonal of 
Then 
can be written in the form 
where 
and 
are positive integers and is not divisible by the square of any prime. What is 
Solution
Let 
denote the intersection point of the diagonals 
and 
. Remark that by symmetry is the midpoint of both 
and 
, so 
and 
. Now note that since 
, quadrilateral 
is cyclic, and so
![\[XR\cdot XA = XP\cdot XB,\]](http://latex.artofproblemsolving.com/5/e/4/5e457e2176d3718620628f1f05e29a00f6821b25.png)
which implies 
.
Thus let 
be such that 
and 
. Then Pythagorean Theorem on 
yields 
, and so![\[[ABCD] = 2[ABD] = AP\cdot BD = 3\sqrt{x^2-1}\cdot 8x = 24x\sqrt{x^2-1}.\]](http://latex.artofproblemsolving.com/2/d/6/2d6ab35c60d4c601a70be2ac18760dd868e35771.png)
Solving this for 
yields 
, and so![\[(8x)^2 = 64x^2 = 64\left(\tfrac12 + \tfrac{\sqrt{41}}8\right) = 32 + 8\sqrt{41}.\]](http://latex.artofproblemsolving.com/e/3/3/e3304ebd189c51dd0bacea140ddf91ce32598c3e.png)
The requested answer is 
.
See Also
| 2021 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
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