Well, nobody ventured forth an answer, and that’s okay– I did preface the problem by saying none of my students was able to solve this. And there is good reason–this problem is close to the 800 level. Unless you are shooting for that score you might not even see a problem that is this difficult.

That said, let’s try to take the problem apart. Here it is again:

The sides of a rectangular box when added together equal 36.

A

The length of the longest possible

line that can be drawn to connect

two of the vertices of rectangular box.

B

8

The first step is to find out how many sides make up a cube. Let’s imagine the face of one the sides and further imagine that we are constructing that side from toothpicks. We will need four toothpicks (a square) for one side. Now, let’s visualize the opposite side. It too will need four toothpicks. Now we want to connect these two faces, by joining the vertices. Each side has four vertices so we will need one toothpick to connect each vertex. That means we will need four toothpicks and a total of 12 for the entire cube.

Now that’s we’ve constructed the cube out of toothpicks, let’s go back to the original problem that asks us to consider a box. The number of toothpicks will not change but the length of toothpicks can now differ. The constraint here is that when you add up the 12 toothpicks they must equal 36. If we kept our initial cube each toothpick would be three inches long. Therefore a cube with side 3, will yield 3√3 for the longest possible line joining vertices (to find this take s, the side of the cube, and multiply a √3.) This gives us a length of approximately 5.1.

With that in mind, let’s now shift our attention to the context of quantitative comparison. When we find the answer to one condition– in this case, if the box is a cube the length of the longest possible line is 5.1. The answer then is (B) 8 , which we will want to disprove. Often times the best way to do so is to think of extreme figures. In a cube each side is equal, so what we are going to do is stretch some sides and shorten others, without ever going over a total of 36. One possible scenario that leaps to mind is 1:1:7, or two tiny squares joined by a 4, length 7 toothpicks. If we add up the length of all the sides we et 36 (you will have to play around with numbers to see which ones fit the conditions of the problem.) Using the Pythagorean Theorem twice–a necessary step to find the longest length of vertices when not dealing with a cube–give us a measurement of approximately 7.1 Not long enough.

Many stop here and say the answer is (B). And that’s why this is such a difficult problem, because even if you get this far you still haven’t gotten the right answer. Nowhere did the question state that the sides had to be integers. Now imagine sides 1/2 : 1/2 : 8, or two extremely small squares joined by length 8. Without having to do any math, we know that the hypotenuse is always longer than the length of the longest side. So therefore the longest side is more than 8. Now a possible answer is (A). We can’t have two possible answers so the answer is (D).

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