Tuesday, December 06, 2011

this week's MGRE Math Beast Challenge

From here:

As usual, my answer will appear in the comments.


1 comment:

Kevin Kim said...

Based on the figure, a right triangle, we're comparing the quantities abc and h(a^2 + b^2). Right away, thanks to Uncle Pythagoras, we know that the parenthetical expression can be converted to c^2.

Let's set up a "mysterious" inequality in which the letter Q represents "greater than, less than, or equal to." The first step of this mysterious inequality would be:

abc Q h(c^2) [based on our deductions]

We can then divide by c on both sides, giving us

ab Q ch

The length h is perforce less than either a or b. The length c, meanwhile, is perforce greater than a or b, but not necessarily greater than ab.


Without actually looking up the properties of right triangles, I'd have trouble seeing how this problem could be solved without more information. So unless someone can swoop in with a better notion, I'm leaning toward (D): the relationship can't be determined from the information given.

But let's try something. Let's divide c into subsections x and y. Let's further assume a and b are equal, thus making x and y equal to each other. Length c can, in such a situation, be rewritten as 2x (or 2y). In this situation, h is at its maximal length relative to the legs and the hypotenuse, and we have a 45-45-90 triangle.

C can be rewritten as a√2.
X and Y can each be rewritten as (a√2)/2.
H can also be rewritten as (a√2)/2, because a 45-45-90 triangle is half of a square. The altitude becomes half a diagonal, just as X and Y are halves of a diagonal.


abc = a*a*(a√2) =



h(a^2 + b^2) =

h(c^2) = [(a√2)/2]*[(a√2)^2] =

[(a√2)/2]*(2a^2) =


Notice that, in a 45-45-90 situation, quantities A and B of the original problem are equal. But since we don't know whether the triangle we're looking at is that kind of triangle, we don't have the information we need to determine the answer. If lengths a and b are unequal, Quantities A and B will be unequal.

So I'm sticking with (D).