## Monday, April 2, 2012

### the answer to last week's MGRE Math Beast Challenge

I blogged last week's MGRE Math Beast Challenge here. Was the answer indeed 4√5, as I had guessed?

Yes, it was. MGRE says:

A rhombus has four equal sides. In this figure, each rhombus side lies on a face of the cube. In order for the rhombus side lengths to be equal, the two corners of the rhombus that lie on a cube edge must lie exactly at the mid-point of that cube edge, as shown:

By Pythagorean Theorem, each rhombus side length is thus √(12 + 22) = √5. If you don’t quite believe that the rhombus corners we placed at the midpoint of the cube edge MUST be at that midpoint, try placing those corners elsewhere on the edge. If the front right rhombus corner lay, say, 1.5 up from the base of the cube and 0.5 down from the top surface of the cube, the front edge of the shaded shape would be √(1.52 + 22) = √6.25, while the right edge would be √(0.52 + 22) = √4.25, i.e., the shaded shape would not be a rhombus.

If each rhombus side length is √5, then the perimeter of the rhombus is 4√5.

We could also apply some common sense and use the answer choices. Each rhombus side length is longer than the edge of the cube, but less than the diagonal of a face of the cube. If x is the rhombus side length, then 2 < x < 2√2. The perimeter of the rhombus must be between 8 and 8√2, exclusive. Approximating √2 as 1.4, the perimeter must be between 8 and 11.2, exclusive.

(A) 8. TOO SMALL
(B) 2√6. TOO SMALL
(C) 4√3. TOO SMALL
(D) 4√5 is between 4√4 and 4√6, which is double choice (B). Between 8 and 9.52. OK
(E) 4√2 + 4√3 ≈ 4(1.4) + 4(1.7) = 12.4. TOO LARGE

I find that that second method takes too long without the use of a calculator.

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