So a Circle Walks into a Bar…


It sounds like the beginning of a math joke, but it isn’t.

“So a right triangle is inscribed into a circle…”

That’s the premise of a couple interesting GMAT questions that I came across lately, so I thought I’d share the issues that these problems bring. First it’s important to define that term inscribed. It’s the kind of term that you may have come across several times without ever knowing what it means because the visual diagram that accompanies the problem has you covered. In geometry, when we talk about something being inscribed we mean that it is drawn inside another shape such that all of its corners touch the edge of the larger shape without going outside of it. When a shape is inscribed within a circle it’s a little like that shape has a custom-built bubble surrounding it.

Now back to our problem. So there’s a right triangle in a bubble. So What? Well that particular situation actually gives us a very important piece of information. Whenever a right triangle is inscribed in a circle, the hypotenuse of the triangle is the diameter of the circle. That’s a fantastic rule, and one you ought to remember, but when we get to the difficult end of the quant section where a question like this is likely to occur, we’re probably going to need more than that.

So what other concepts fit in with this rule? Well, our rule gives us a fantastic way to find the hypotenuse of the triangle if we know something about the circle (or vice-versa), so a nice extra step is when the GMAT asks about the length of one of the other sides of the triangle. When would we be able to find the length of the other sides of the right triangle knowing only the length of the hypotenuse? When it’s a special right triangle! So, be on the lookout for 30:60:90 triangles or 45:45:90 triangles. Even if these aren’t immediately apparent, remember that every distance from the center of the circle to the edge of the circle is a radius, and drawing one or more of these radii in often gives you more opportunity to solve.

Keep this fantastic rule and these tips in mind the next time you come across a similar problem!



I recently came across a problem that required knowledge of parabola formulas. That surprised me. Although parabola problems show up relatively frequently, they usually require little more than logic. However, I thought that provided a nice opportunity to refresh on all things parabola.

A parabola is defined mathematically by this formula: y= ax^2 + bx + c. We see parabolas in nature most often when we look at projectiles, like a cannonball shot out of a cannon or a jump shot out of the hand of Steph Curry. Parabolas are generally u-shaped and are symmetrical about the vertex, which is either the highest or lowest point of the parabola, depending on the orientation.

Whether our parabola is cupped upward or downward is determined by the sign of the “a” term in the formula we saw above. When a is positive, the parabola will have a vertex at the bottom and open upward. When a is negative, the parabola will have a vertex at the top and open downward.

However, there is more than one way to define a parabola mathematically. We can also solve a parabola if we have the vertex and another point on the parabola. We do that by using the similar formula y = a(x-h)^2 + k. The coordinates of the vertex are (h,k).

So using that information, find the equation of a parabola with vertex (-2,1) containing the point (1, 19). The first thing we need to do is solve for a by inserting our points into the formula. We get:

19 = a(1-(-2)^2 + 1

19= a(3^2) + 1

19= 9a + 1

Now, we put our a into the formula with our vertex (h,k), but instead of using the x and y from a specific

point we’re going to solve for the generic x and y.

y= 2(x-(-2)^2 + 1

y= 2(x+2)^2 +1

Now expand:

y=2(x+2)(x+2) +1

y=2x^2 + 8x + 9

Now we’ve solved for the equation of this parabola and we could mathematically figure out all of the

points on this curve.

I hope that’s been a good refresher on parabolas!

 Change in Christian’s Pocket


Christian has 13 coins in his pocket. The total value of the change is $1.37. The change may be composed of pennies, nickels, dimes and quarters. How many dimes could be in Christian’s pocket?

A) 0

B) 3

C) 5

D) 7

E) 9

There are several ways to approach this problem. Many of them are wrong. That’s because the core problem (13 coins, $1.37) can be solves several different ways. If you approached this and just decided to solve it you might find two or three solutions before you found one that was in the answer choices. That’s the first important step. When you’re given a problem without a clear, obvious and quick solution look to the answer choices.

Once, we’ve decided that we’re going to work through the answer choices, there is still some uncertainty. Look at answer choice B for instance. Even if we know that we have 3 dimes, that still doesn’t sort out how many pennies, nickels and quarters we have. That brings us to tip number two: when you’re given complex information, look for opportunities to simplify. The best example of that in this case is the pennies. Because our total needs to end in a 7, and all the other coins can only get us multiples of 5, we know that we must have either 2 or 7 pennies in order to make this work. That makes quick work of checking the possibilities for answer choice B. If we have 3 dimes and 2 pennies, can we make $1.05 out of 8 coins? Well, quarters can only get us to multiples of 25, so we must have at least one nickel so we can simplify again. Can we make $1.00 composed of quarters and nickels? Nope. Doesn’t work. Test 3 dimes and 7 pennies and you’ll very quickly see that you can’t make $1.00 out of 3 of the coins we’re given.

The final tip is that challenging problems don’t need to be tough. If you went into this and said I’m not sure how to solve this, but I can work from the answer choices and I can simplify problems, you might have started by testing answer A.

If I don’t have any dimes, I might have exactly 2 pennies. If that’s the case I have 11 coins to make $1.35. I need at least 2 nickels, so now I have 9 coins to make $1.25. 4 quarters and 5 nickels would work. Done. Answer choice A is correct. (You can also make this work with 0 dimes and 7 pennies).

When you don’t know where to start, look at the answer choices and look to simplify.

7 pennies, 1 nickel, 0 dimes, 5 quarters

 Now and Venn


One of the books I’m currently reading is Priceless: The Myth of Fair Value (and How to Take Advantage of It) by William Poundstone. It’s an interesting look as some of the psychology behind how we think about prices and perceive value. Poundstone recounts one experiment done by Kahneman and Tversky called “Linda the Bank Teller” that I’d like you to think about.

“Linda is 31 years old, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations.”

Which of the following is more likely to be true?

Linda is a bank teller.

Linda is a bank teller and is active in the feminist movement.

Astonishingly, 85% of the college students tested responded that the second statement is more likely to be true than the first, which is absolutely ridiculous. The point that Kahneman and Tversky were making is that we rely on heuristics or “rules of thumb” more often than is logically sound. The point I’d like to make is about drawing diagrams.

The information you’re given in this situation, although different in format from what you’re going to see on a standardized test is nonetheless similar. You’re given a few different pieces of information that fit together in some way that may not be intuitively obvious. One of the best ways to deal with that situation is to draw a graphical representation of what you’ve been told. When given two overlapping sets, you should draw a Venn diagram.

Now, let’s consider the two situations. First, the chance that Linda is a bank teller is represented by the region in red.

The chance that Linda is a bank teller and is active in the feminist movement is represented by the region in blue.

As you can see, the second set is a subset of the first. A drawing with only the two regions we’re concerned with would look something like this:

Linda can’t be a feminist and bank teller without being a bank teller. In this graphic form it’s pretty obvious, and yet 85% of the students missed this fact the first time around. This is a great example of the benefits of converting words into graphical forms so that you can make sure you have a good understanding of what you’re doing before you go forward with a problem.