Varsity Math 85


This week, Coach Taylor tries her hand at a pair of puzzles on the theme of right triangles.




Coach Taylor calls a right triangle “perimetric” if all of its sides have integer lengths, measured in the same unit, and its perimeter is numerically equal to its area (measured in square units of the same unit as the side lengths were measured).

Find all perimetric right triangles.

Hypotenuse Partition

The largest possible semicircular arc S is inscribed in the right triangle R so that the center of S is on the hypotenuse of R. The two endpoints of S and its center split the hypotenuse into four segments. The lengths of three consecutive segments out of those four, in order from one vertex of the hypotenuse toward the other, are 16, 120, and 120.

What is the length of the fourth segment (which has as one of its endpoints the other vertex of the hypotenuse)?


Solutions to week 84

Cucina Combinations. First, let’s figure out how many different pizzas you could order, depending on the number of toppings, t. There’s just one pizza with no toppings and t pizzas with one topping. For two toppings, there are t ways to choose the first topping, but only t-1 ways to choose the second topping. And since the toppings get mixed together on the pizza, the order you choose the two toppings doesn’t matter, so we need to divide by two to get the number of different pizzas with two toppings. So there are t(t-1)/2 pizzas with two toppings. Similarly, for the pizzas with three toppings, you have t(t-1)(t-2) ways to list three toppings, but you have to divide by six because there are six different orders you could list three toppings, but the order of the toppings doesn’t matter. Adding it all up, there are t(t-1)(t-2)/6 + t(t-1)/2 + t + 1 different pizzas with three toppings, which we could multiply out as (t³ – 3t² + 2t)/6 + (t² – t)/2 + t + 1 and then add up to get (t³ + 5t + 6)/6 different pizzas.

On the other hand, for the lasagna, there are t+1 choices for the bottom layer (any topping or no topping) and t+1 choices for the top layer, and here order does matter, because a lasagna with spinach on the bottom and sausage on top is different from one with sausage on the bottom and spinach on top. So there are (t+1)(t+1) = t² + 2t + 1 different lasagnas.

What Riley tells us is that the number of pizzas minus the number of lasagnas is 1,265, or in other words, (t³ + 5t + 6)/6 – t² – 2t – 1 = 1265. Combining terms on the left, and multiplying both sides by six, we get that t³ – 6t² – 7t = 7590. Note that the left hand side is divisible by t, and in fact it factors as t(t-7)(t+1) = 7590, so we just look at the list of factors of 7590, i.e., 1, 2, 3, 5, 6, 10, 11, 15, 22, 23, 33, 46, 55, 66, … to find a factor so that seven less is also a factor and one more is also a factor. The first possibility is ten, but 10×3×11 is too small. The next possibility is 22, which turns out to work: 22×15×23 is indeed 7590. So the restaurant offers 22 different toppings.

Potato Puzzle. Suppose Avery’s habit started d days ago. Then d days ago, Avery ate 1 chip; d-1 days ago, Avery ate 1+m chips; d-2 days ago, Avery ate 1+2m chips, and so on, until yesterday Avery ate 1+(d-1)m chips. The average number of chips eaten per day is therefore (2+(d-1)m)/2 chips, so the total number of chips Avery has eaten (namely, 1177) must equal d(1 + (d-1)m/2) chips.

So if d is odd, then 1 + (d-1)m/2 is a whole number, and d is a factor of 1177; whereas if d is even, then 1 + (d-1)m/2 might be half an odd number, so we can conclude that either d or d/2 is a factor of 1177. Also, since 1 + (d-1)m/2 is bigger than one, d or d/2 must be a proper factor of 1177. However, 1177 = 11×107, both of which are prime, so we conclude that d must be 11,22,107, or 214. We try these in order: if d were 11, then 1 + (d-1)m/2 = 107, or 10m = 212, which is impossible. If d were 22, then 1 + (d-1)m/2 = 107/2, so 21m = 105, with the solution m=5. If d were 107, then 1 + (d-1)m/2 = 11, or 106m = 20, which is impossible, and finally if d were 214, then 1 + (d-1)m/2 = 11/2, or 213m = 9, which is impossible. Hence, we conclude that Avery’s habit started 22 days ago.

Coach Newton can’t resist saying:
How do you like that? I gave you two puzzles, and the answer to each of them was the day of the month that we held our practice session. Not bad, eh?

Recent Weeks

Week 84: Potato Puzzle & Cucina Combinations, solutions to Sour Lemons & One Liner

Week 83: Sour Lemons & One Liner, solutions to Curious Clock & Quadratic Triple

Week 82: Curious Clock & Quadratic Triple, solutions to Poor Pascal & Rascal’s Grid

Week 81: Poor Pascal & Rascal’s Grid, solutions to Secret Remainder & Party Puzzle

Week 80: Secret Remainder & Party Puzzle, solutions to Wing Nuts & Seven Eleven

Links to all of the puzzles and solutions are on the Complete Varsity Math page.

Come back next week for answers and more puzzles.