Summer Problem Solving Marathon Solution #3

From a + d = b + c, we have a = b + c - d. We substitute this for a in bc - ad = 93 to get:

bc - (b + c - d)d = 93

or:

bc - bd - cd + d² = 93

Rearranging terms, we get:

bc - cd + d² - bd = 93

Factoring out c from the first two terms and d from the second two terms, we have:

c(b - d) + d(d - b) = 93

or:

d(d - b) - c(d - b) = 93

Then factoring out d - b, we have:

(d - c)(d - b) = 93

Now d - b and d - c must be integers, and d - b > d - c (since c > b). Since 93 = 3 * 31, there are two possibilities:

#1: d - b = 93, d - c = 1.
#2: d - b = 31, d - c = 3.

On the first possibility, b = d - 93, and c = d - 1. Since a = b + c - d, we have a = d - 93 + d - 1 - d = d - 94.

Since d < 500, the greatest possible value for d is 499. Since a = d - 94 > 0, the smallest possible value for d is 95. Once we pick d, the values of a, b, and c are fixed. There are 499 - 95 + 1 = 405 possible ordered 4-tuples under this possibility.

On the second possibility, b = d - 31 and c = d - 3. So a = b + c - d = d - 31 + d - 3 - d = d - 34.

So d is at most 499 as before, and at smallest 35 (so that a is at least 1). So there are 499 - 35 + 1 = 465 possible value for d on this possibility, and hence 465 possible ordered 4-tuples.

So between the two possibilities, there are 405 + 465 = 870 possible ordered 4-tuples meeting the conditions.

Summer Problem Solving Marathon Solution #2

The perpendicular from A to BC creates two 30-60-90 right triangles:



Because the two triangles ABD and ACD are congruent, BD and CD must both be 6:



In a 30-60-90 right triangle, the sides are in length ratios of 1 - √3 - 2. In the 30-60-90 right triangle ABD, the three sides in this ratio are BD, AD, and AB. Since BD is of length 6, AD must be of length 6√3 and AB of length 6*2 = 12 (which it is):



We now add point E at the midpoint of AD:



Since AD is of length 6√3 and E is its midpoint, ED is of length 3√3.

Now we add the line BE, which we want to find the length of:



This creates the right triangle BDE. We know that BD is of length 6, and DE of length 3√3. Using the Pythagorean Theorem, we then have:

BE =
√ 6^2 + (3√3)^2 

=
√ 36+27 

=
√ 63 

=3√7

Summer Problem Solving Marathon Question #4

[Value = 4 points]

In the expansion of (x - ¹⁄_√x)⁷, what is the coefficient of x^-1/2?

Summer Problem Solving Marathon Question #3

[Value = 7 points]

How many ordered four-tuples of integers (a,b,c,d) are there such that:
(i) 0 < a < b < c < d < 500
(ii) a + d = b + c
(iii) bc - ad = 93

Summer Problem Solving Marathon Question #2

[Value = 3 points]

Each side of triangle ABC is length 12. D is the foot of the perpendicular dropped from A on BC, and E is the midpoint of AD. What is the length of BE?

Summer Problem Solving Marathon Solution #1

Let x and y be the two numbers. Then we are told x + y = 10, and xy = 20. We want to find ¹⁄_x + ¹⁄_y.

Converting both fractions to the common denominator xy, we get ^y⁄_xy + ^x⁄_xy, or ^(x+y)⁄_xy.

Substituting our starting values for x + y and xy, we have ¹⁰⁄₂₀, or ¹⁄₂.

[Put any questions, objections, or alternative solutions in the comments.]

Summer Problem Solving Marathon Question #1

[Value = 2 points]

The sum of two numbers is ten, and their product is twenty. What is the sum of their reciprocals?