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!

 The New PSAT: Math Quick Thoughts


I recently had an opportunity to take a sample test for the new PSAT, and I have to say I was surprised. For all the bluster about a new SAT I didn’t expect a whole lot of difference. I will say, it’s different. Here are some of my high-level thoughts about the math section. I’ll be digging into more details of the test and more specifics in coming weeks.

1. The new non-calculator section is going to be a major time crunch: I’m traditionally a fast worker, I’m extremely comfortable doing mental math and I still only finished with about a minute and a half left.

2. Finishing on time is going to require either exceptional math skills or exceptional strategy: And to be honest, a combination is probably the best path to a great score. One problem I saw involved some fairly complicated geometry. However, a quick estimate told me that only one of the answers could have been correct and saved me quite a bit of time. Mindlessly plugging ahead is a sure way to get in time trouble on this test.

3. The combination of time pressure and some tough problem should make perfection tougher to achieve: If you’re not an 800-level math student, that shouldn’t cause you to panic because the scoring will still curve out approximately the same. My guess is that the need for a little more speed may do a better job separately the truly exceptional from the very very good. The current test often makes silly mistakes the dividing line between perfect and near perfect.

I’m still digesting what these changes mean, but I’ll keep updating you over the coming weeks!

 Closing the Gap


Chris and Evan are outstanding runners. One day in practice they do a 1500 meter training run. Being competitors, Chris and Evan can’t do any training run without it turning into a competition, so they decide to race. Chris and Evan were dead even until Chris tripped at the 1100 meter mark. His misfortune allowed Evan to take a 15 meter lead. Jolted by adrenaline Chris immediately ups his pace to 6.7 meters/second. Evan continues at his consistent pace of 6.4 meters/second. Who wins the race?

This problem mimics some themes we often see in math problems in content, although not exactly in form. The common thread is that you have two rates. Dealing with two rates simultaneously, however, can be tricky, so it’s easier to deal only with the difference between the rates. In this case (at least at first), let’s not worry about 6.4 and 6.7. All we need to know is that Chris gains 0.3 meter every second. With a difference of 15 meters, it’s quick to figure out that it will take 15/0.3 = 50 seconds to close the gap.

Only one question remains: When Chris does catch up, will it be before or after the finish line? Since we know the two runners will be tied 50 seconds after the trip, all we need to do is multiply one of the runners’ rates by 50 seconds to figure out where on the course he will be when that occurs. Let’s chooses Chris. 6.4 m/s * 50 seconds = 320 meters. Adding that to the 1115 meters he has already traveled (since he’s already gained the 15 meter lead by the time both men are running again) and we find that Chris will be 1435 meters into the race when Evan pulls even with him.

Despite his trip, Evan will pull even with Chris and pass him down the back stretch to win the 1500 meter race.

 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