Three of the vertices of a square in ℝ³ are (1, 2, 3), (10, 14, 23), and (22, 30, 8).  What is the sum of the coordinates of the fourth vertex?

Solution: The square in question is the translation by (1, 2, 3) of the square with vertices (0, 0, 0), (9, 12, 20), (21, 28, 5), and (12, 16, 15), so the fourth vertex is
(12+1, 16+2, 15+3)=(13, 18, 18), whose sum of coordinates is 49.

Two circles are concentric. A chord c units long cuts across the larger, tangent to the smaller. What is the area of the shaded region in terms of c?

Solution: [From MT Nov 1991]

Let the radii of the smaller and larger circles be r and R respectively, so the area we seek is given by ℼR² – ℼr².  Now, let A be the center of both circles, let point B be the point of tangency of the chord, and D be the point of intersection with the larger circle, as indicated.  Now, because triangle ABD is right, we have r² + (c/2)² = R², so that R² – r² = (c/2)².  The shaded area is then equal to ℼc²/4.

The integers 1 thru 100 are written on individual cards, and those 100 cards are mixed into a hat. Aisha draws one card at a time, without replacement, until she draws a card relatively prime to all the cards she has already collected. What is the maximum number of cards Aisha could draw in this game?

Solution: 90.  One can construct a sequence of maximal length as follows:

• Starting with 2, include all the even numbers from 2 through 100, a total of 50.
• Ignoring 1, now include all the odd numbers that are not primes over 50. Since there are 10 primes between 50 and 100, there are 49-10=39 of these.
• The 90^th pick will now either be a 1 or a prime over 50, all of which will be relatively prime to all those on the list.

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).

The integer a ends in a string of n zeros. If

what is n?

Solution:

so a = 100!.  The number of zeros at the end of 100! is the same as the number of 5s in its prime factorization. There are 20 factors of 5 in the first 100 integers, but 4 of them are also factors of 25, yielding 24 factors of 5 in all.

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.

Three integers a, b and c satisfy both 0 ≤ a ≤ b ≤ c and

          abc + ab + ac + bc + a + b + c = 2023

What is the smallest possible value for a + b + c?

Solution: Add 1 to both sides and factor to get (a + 1)(b + 1)(c + 1) = 2024.

We pick a = 7, b = 10, and c = 22, so that a + b + c = 39.

Three integers a, b and c satisfy both 0 ≤ a ≤ b ≤ c and

          abc + ab + ac + bc + a + b + c = 2023

What is the smallest possible value for a + b + c?

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,

In the diagram below, the radius of each of the smaller circles is r.  What is the radius of the larger circle?