AMC 8 2012 Problem 25

Happy 4th of July!! I don’t have a good problem for you today, but let’s tone it down a little with this:

Problem:

[asy] draw((0,2)--(2,2)--(2,0)--(0,0)--cycle); draw((0,0.3)--(0.3,2)--(2,1.7)--(1.7,0)--cycle); label("$a$",(-0.1,0.15)); label("$b$",(-0.1,1.15));[/asy]

A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length $a$, and the other of length $b$. What is the value of $ab$?

$\textbf{(A)}\hspace{.05in}\frac{1}5\qquad\textbf{(B)}\hspace{.05in}\frac{2}5\qquad\textbf{(C)}\hspace{.05in}\frac{1}2\qquad\textbf{(D)}\hspace{.05in}1\qquad\textbf{(E)}\hspace{.05in}4$

My “Attempt”:

Since this is an AMC 8 problem, I probably did this before. Anyway, it seems quite straight-forward. The bigger square has an area of , so it must have a side length of . Therefore, .

Also, the smaller square has an area of , so it must have a side length of . Since all of the “long partitions” of the side of the square is and all the “short partitions” are (because of the ASA triangle congruences):

3d4cdd3dbcabe3dc76b806d60b785416afc65f11

Because of this, . How do we find out of the equations and ? We take the square of the first equation. Now,


Subtracting from that equation, we get , so . Therefore, The answer is C. Let’s see if we’re .

Solution:

Oh man. We’re right, of course, but there’s an easier way to do the problem. According to the solution, the area of one of the small right triangles is . Therefore, . So apparently, I was kind of outsmarted by an AMC 8 problem. Quite the shame.

Conclusion:

This only took about 20 minutes to write, and it’s just a filler problem. I will definitely make a post tomorrow, though!

And the time is .

 

 

 

 

 

 

 

 

 

 

 

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