answer to last week's MGRE Math Beast Challenge

The answer to last week's MGRE Math Beast Challenge was indeed (B). But MGRE shows how to arrive at this answer without using Heron's Formula. To wit:

To find the area of triangle PQR we need a base and a height. If we consider side PR the base of the triangle, then QS, which is at a right angle to PR, is the height. We know that both triangle PQS and triangle SQR are right triangles, but to find the height QS we’ll need to know the length of either PS or SR. Unfortunately we don’t know either one, so we’ll have to name variables for several legs of the triangle.

In this case we’ll let x represent the length of PS and y represent the height QS. Note that if PS has length x, then SR has length 21 – x, so we do not need to (and should not) name a variable for length SR. Updating our diagram yields the following:



Apply the Pythagorean Theorem to each right triangle.
From triangle PQS: x² + y² = 100
From triangle SQR: (21 – x)² + y² = 289

We solve each in terms of y²:
y² = 100 – x²
y² = 289 – (21 – x)²

Then set the two expressions equal to y² equal to each other:
289 – (21 – x)² = 100 – x²

Now simplify:
189 – (21 – x)² = -x²
189 = (21 – x)² – x²
189 = (441 – 42x + x²) – x²
189 = 441 – 42x
42x = 252
x = 6

Now that the value of x is known, solve for y:
100 = 36 + y²
64 = y²
8 = y
(Or just recognize that PQS is a 6 – 8 – 10 right triangle.)

Finally, we can solve for the area of triangle PQR:

(1/2)bh = (1/2)(21)(8) = 84

The correct answer is B.

{Incidentally, this problem is named for Heron’s Formula, which is an alternative to the (1/2)bh triangle area formula. For a triangle with side lengths a, b, and c, the semiperimeter is defined as s = (a + b + c)/2. The area of the triangle equals √(s(s-a)(s-b)(s-c)). Note that this doesn’t require a right triangle, nor does it require knowledge of any heights measured perpendicular to any base. You will NOT need to know this for the GRE, as the original solution above attests.}

I was glad to see the above explanation, because I was honestly stumped as to how to approach the problem. I had thought that Heron's Formula would be the only way to solve it.


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this week's MGRE Math Beast Challenge

From here:



Go to it! My own response will appear in the comments.


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answer to last week's MGRE Math Beast Challenge

The correct answer was indeed (B)!

Here's MGRE's explanation:

This problem gives us an equilateral triangle and a circle, and tells us that the perimeter of the equilateral is 1.25 times the circumference of the circle.

This is license to plug in. Since both an equilateral and a circle are regular figures—that is, all equilaterals are in the same proportion as all other equilaterals, and all circles are in the same proportion as all other circles—we can be certain that we only need to plug in one set of values in order to be sure of the answer. Because we have regular figures and a way to relate them (triangle perimeter = 1.25 × circumference), we will not need to repeatedly try different values as we often do on Quantitative Comparisons.

We could say the radius of the circle is 2, so the circumference is 4π. Then, the perimeter of the equilateral triangle is (1.25)(4π) = 5π . This isn’t ideal, because then we are stuck with π in the calculations for the triangle, where it is unnecessarily awkward.

We could say circumference = 4 and therefore the perimeter of the triangle = (1.25)(4) = 5. But then the side of the equilateral triangle is 5/3, a non-integer, which is inconvenient. We can avoid this by using larger numbers (multiplied by a factor of 3).

So our smartest numbers are circumference = 12 and therefore the perimeter of the triangle = (1.25)(12) = 15, so each side of the triangle is 5. The height of an equilateral triangle is always (√3)/2 the side length. You can always derive this yourself by splitting the equilateral triangle into two 30–60–90 triangles, with known side ratio 1x : (√3)x : 2x. So, the area of the triangle is

(1/2)bh =

(1/2) • (5) • ((5)((√3)/2)) =

(25√3)/4 =

approx. 10.625. (Use the calculator.)

In the circle, circumference = 12 = 2πr, so

r = 12/(2π) = 6/π. The area of the circle is

π(r^2) = π((6/π)^2) = 36/π = approx. 11.46.

(Use the calculator and the approximation 3.14 for π.)

The correct answer is B.

I have to say... I like my method a lot better.


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this week's MGRE Math Beast Challenge

From here:

The perimeter of an equilateral triangle is 1.25 times the circumference of a circle.

Quantity A
The area of the equilateral triangle

Quantity B
The area of the circle

(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.

Go to it! My answer will appear in the comments.

By the way, sorry about the lack of posting last week. I think my sickness lingered on a bit longer than I had thought.


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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.

The correct answer is D.

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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this week's MGRE Math Beast Challenge: blind edition

MGRE's website is undergoing some sort of facelift, which is playing havoc with certain parts of the site. This week's Math Beast Challenge has been affected: I can no longer see the illustration that accompanies the problem, nor can I see most of the answer choices, except for the first choice, which is 8. So this week, we're going to fly blind and just use our imagination to try to draw the illustration accompanying the problem. The problem itself is pretty clearly worded, and we needn't worry about the unseeable answer selections: we'll just figure things out on our own.

A solid cube with edge length 2 is sliced by a plane passing through two opposite corners of the cube. This creates a cut surface, the shaded rhombus shown [below]. What is the perimeter of the shaded rhombus?

I've attempted to make an illustration, which you see below. Note that the illustration is in two parts: a drawing based on the above description, plus a "closeup" that will help us analyze the problem.




I'm of the opinion that the perimeter of the rhombus in question is 4√5. Segment AC is the hypotenuse of a right triangle whose legs measure, respectively, 2 (Segment AB) and 1 (Segment BC). We can check visually that all the other edges of the rhombus are the same length, so this rhombus is a square with perimeter 4√5.

I bet that, once MGRE's site is back in order, one of the answer choices will indeed be 4√5.

By the way, here's a horizontally squished version of what I currently see at the MGRE page:




As you see, there's no illustration, and only one visible answer choice.


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SAT-style geometry problems

A good set of geometry problems for the SAT can be found here. I discovered these while planning a geometry tutoring session for my goddaughter. The Khan Academy website also has its own fairly comprehensive geometry section, as well as a massive SAT Math prep section.

Before I forget: this Cliffs Notes page does a very good job of explaining the relationship between circles, chords, secants, and tangents.


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this week's MGRE Math Beast Challenge problem

From here:




My answer will appear in the comments, but I'll tell you right now that I don't like the way this problem's been structured. As it's worded, I'd say the answer is "cannot be determined." Why? A couple reasons:

1. We can assume the figure isn't drawn to scale. The only "given" is that the round figure is indeed a circle with its center labeled. We can also safely assume that Points A, C, and D are all on the circle. It's also safe to assume that Point B is on the circle as well. Beyond that, what do we know for sure?

2. We can't assume-- since the problem doesn't specify this-- that the squarish-looking figure is indeed a square, which means we can't assume that the circle is properly inscribed within a square.

3. We also can't assume that Segment GF forms a 180-degree angle with the bottom of the putative square.

For those reasons, I think "cannot be determined" is the best answer, but I'm going to ignore the above concerns and charitably assume that the circle is inscribed in a square, thus making Point B the midpoint of that side of the square. With those assumptions in place, I believe the problem is easily soluble. Without them, however, "cannot be determined" is the only legitimate answer.

Here's how I would have presented the problem, so as to avoid any confusion about what we can and can't assume:



I'll be basing my answer off the above image, not off MGRE's.


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this week's MGRE Math Beast Challenge problem

From here:

This Week's Problem: "Honeycomb"

The honeycomb figure above consists of six identical regular hexagons, each with side length 2√3.

Quantity A
The total area of the honeycomb

Quantity B
180

(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.

Good luck! My own answer will appear in the comments. (Please show your work if you decide to submit your own answer!)


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this week's MGRE Math Beast Challenge problem

From here:

"Walter's Exercise"

Every day, Walter burns 500 calories from cardio exercise. On some days, he also burns an additional 600 calories from weight training. If, over a 240-day period, Walter burns an average of 850 calories per day from cardio exercise and weight training combined, then on how many more days did Walter engage in both cardio exercise and weight training than in cardio exercise only?

(A) 40
(B) 60
(C) 80
(D) 100
(E) 140

My answer will appear in the comments.


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MGRE problem: wrong!

Even a teacher can get things wrong, and this time around, I did.

My answer to last week's Math Beast Challenge problem turns out to be incorrect. You'll recall that my answer was (D); MGRE's answer is (C): the quantities are equal. And they're right. But why? Because of one little fact about right triangles that I had missed: the triangle's altitude, drawn from the vertex of the two legs to a point on the hypotenuse, creates two right triangles that are geometrically similar to the large triangle. I should have realized this. Anyway, without further ado, here's part of MGRE's explanation for why (C) is correct:

This could be solved with the Pythagorean Theorem, as there are three right triangles in the figure: the small one on the left, the bigger one on the right, and the largest right triangle comprised of the other two. It should also be noted that these three triangles are similar triangles; that is, the three triangles have the same three angle measures.

For the largest triangle, a² + b² = c² so by substitution, Quantity B = hc². Now that Quantity B is more similar in form to Quantity A, we will compare.

Quantity A: abc
Quantity B: hc²

Divide both quantities by c. Dividing both quantities by the same positive number will not change the relative values; the larger quantity will still be larger. This comparison becomes

Quantity A: ab
Quantity B: hc

For similar triangles, the ratios of side lengths will be equal. For example, the ratio of the short leg to the hypotenuse will be the same in each triangle.

(short leg)/(hypotenuse) = a/c (from the largest triangle) = h/b (from the triangle on the right)

a/c = h/b

By cross-multiplying, we conclude that ab = hc and thus the two quantities are equal.

MGRE's explanation continues, but it's basically a plug-in-the-numbers approach. What bugs me is that I was obviously on the right track, but I stopped in my ruminations before I'd figured out the "similar right triangles" part. Had I done that, I'd have seen that the equality I had discovered for one case (45-45-90 triangles) must also obtain for all cases.

Live and learn, eh?


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