Are You Smart Enough To Find The Length?
I was doing this geometry problem with my friend as a contest to see who would get to the answer faster. After trying for 30 minutes, we gave up.
I knew I was so close to the answer but my method was not quite there.
Instead of focusing on the geometry alone, a quicker way would be to make use of coordinate geometry. And that’s the biggest hint to this puzzle.
Here’s what we will do. We denote the side of the square to be S, and the radius of the big circle to be R. Naturally S = 2R. The key to the solution lies in placing the figure onto the xy-plane. Here the bottom left of the square is at the origin (0, 0). And the center of the circle has coordinates (R, R). Let’s zoom in on the top left corner. There’s quite a bit to unpack here. Remember that the radius of each of the 12 small circles is 1. This means the center of the right circle in the top left corner must have x-coordinate 1 + 1 + 1 = 3…
✓ Solution
Here’s what we will do. We denote the side of the square to be S, and the radius of the big circle to be R. Naturally S = 2R.
The key to the solution lies in placing the figure onto the xy-plane.
Here the bottom left of the square is at the origin (0, 0). And the center of the circle has coordinates (R, R).
Let’s zoom in on the top left corner.
There’s quite a bit to unpack here. Remember that the radius of each of the 12 small circles is 1.
This means the center of the right circle in the top left corner must have x-coordinate 1 + 1 + 1 = 3, and y-coordinate S − 1. Remember S is the side length of the square.
Since we know S = 2R, we now have two points in terms of R, and we also know its distance in terms of R.
That’s good news because we can just apply the distance formula! Woohoo!
Upon expanding and simplifying, we should get the following quadratic equation.
And this has two roots.
Either R = 9 or R = 1. But R must be greater than 1 so we find the radius to be 9.
And so the side of the square must be 18.
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