Solution to Mathematical Toy Box – AQ2024 #1

The coefficients of the quadratic function f(x)=ax2+bx+c are chosen from the numbers 1 through 6 by three independent rolls of a fair die. What is the probability that the parabolic graph of  y=f(x) has its vertex on the x-axis?

Solution:  5/216 . There are  63=216 possible outcomes for the ordered triple of coefficients (a, b, c). The vertex of the graph will lie on the x-axis if and only if the discriminant b2-4ac=0. Since b must therefore be even, the possibilities can be narrowed down quickly to these five:

(1, 2, 1), (2, 4, 2), (1, 4, 4), (4, 4, 1), and (3, 6, 3).

Solution to Mathematical Toy Box – SQ2024 #1

There are three houses, equally spaced on a street, labeled in order A, B, and C, in which five children live. There is at least one child in each house. The children decide on a weekend to arrange a playdate, and the meeting place is the house that minimizes the total distance traveled. What is the probability that the children meet at house B?

Solution: One can track all the possible configurations of children distributed in the three houses, along with the total distance (number of segments) travelled by the children in attending the playdate at the various houses:

We observe that the minimum total distance occurs in house B in 4 out of the 6 cases (highlighted blue), giving a probability of 2/3.

This problem is based on a 1950s Putnam problem, which involved a much more general result. For a shortcut, we note that the house that minimizes the total distance is the one in which where Child #3 resides.

Solution to Toy Box – WQ2024 #1

Let A, B, and C be the centers of the smaller circles, let E be the point of tangency of circles B and C, and let D be the center of the large circle, as indicated in the figure below.

Then triangle ABC is an equilateral triangle with side lengths 2r, and triangle CED is a right triangle with length of CE equal to r.  Further, because triangle ABC is equilateral, angle Θ is half of angle ACB, or π/6 .  If R is the radius of the large circle, then clearly

Now,

so that

and therefore,

Hence,

Solution to Toy Box – Fall 2023

Problem: If all possible (distinguishably different) arrangements of the letters A, C, C, L, L, S, U, and U are put in alphabetical order, in what position is the word “CALCULUS”?


Solution: The total number of distinguishably different arrangements of the letters A, C, C, L, L, S, U, and U is given by

To find the appearance of “CALCULUS,” we first, we need to “clear” all the arrangements that begin with A, of which there are

Those arrangements that begin with “CAC” will appear next on the list and need to be cleared as well. There are

of these, the number of distinguishable arrangements of the remaining letters L, L, S, U, and U.

The next several entries on the list will begin with “CALC,” followed by the letters L, S, U, and U.  We can first clear the arrangements that start with “CALCL,” numbering

and then those that begin “CALCS,” which also number

Then the next arrangement on the list is “CALCULSU,” followed immediately by “CALCULUS.”  Hence, “CALCULUS” appears in position number
630 + 30 + 3 + 3 + 1 + 1 = 668.