blackpanther

First we fix the psychology book in the middle. There are 3 choices we can make for this book.

For the remaining 6 slots (3 on either side of the psychology book), we can put any of the 7 textbooks in the first slot.

Once we put a book in the first slot, we have 6 choices left for the second, 5 choices left for the third, and so on.

So, following the fundamental counting principle, we might be tempted to say there are 7⋅6⋅5⋅4⋅3⋅2=(7−6)!7!​=5040 ways of arranging the books.

However, this overcounts all the arrangements where we just move all the math books to one side and all the psychology books to the other side.

There are 2! ways of doing this (since we can switch the two sides), so we have to divide our previous answer by 2.

Therefore, there are 7!/(7- 6)!​ * 1/2! = 2520​ ways of arranging the textbooks.

We can solve this problem by utilizing the properties of medians and similar triangles in a trapezoid.

Median Property: A median of a trapezoid divides the bases into segments with equal lengths. In this case, since MN is a median, we have:

PQ=MQ+QP

RS=SR+RN

Similar Triangles: Since MN is a median, it intersects diagonals PR and QS at their midpoints (X and Y respectively).

This creates two pairs of similar triangles:

Triangle MPX is similar to Triangle NQY (due to AA similarity - corresponding angles at P and Q are congruent since they are alternate interior angles of parallel lines PQ and RS, and shared angle at M/N)

Triangle MRX is similar to Triangle NSX (due to AA similarity - corresponding angles at R and S are congruent since they are alternate interior angles, and shared angle at X)

Leveraging Similarity: Since triangles MPX and NQY are similar, we have the following proportion:

NQMP​=NYMX​

Similarly, from triangles MRX and NSX:

NSMR​=NXMX​

Relating Segments: We are given that XY = 5, which translates to MX + NY = 5. Since X is the midpoint of PR, we know MX = XR and similarly, NY = YN. Substituting these into the first equation from step 3:

NQMP​=NYMX​=YNXR​

Similarly, since X is the midpoint of QS, we know MX = XS and similarly, NX = NY. Substituting these into the second equation from step 3:

NSMR​=NXMX​=NYXS​

Connecting to Bases: We know that RS = 28, which can be further divided using the median property:

RS=SR+RN=MR+NS+RN=MR+NS

Combining Information: We now have two equations relating the ratios of segments on the bases (MP/NQ and MR/NS) to the segments along the diagonals (XR/YN and XS/NY). We also have an expression for RS in terms of segments along the median (MR + NS).

Since we are looking for PQ, we can rewrite it using the median property:

PQ=MQ+QP=NQ+MR

Solving for PQ: To solve for PQ, we can try to eliminate terms involving segments along the diagonals (XR, YN, XS) from the equation for PQ.

One way to achieve this is to notice that the ratios XR/YN and XS/NY appear in both equations we derived in step 5. If we can express one ratio in terms of the other, we can eliminate it.

From the equation for RS:

RS=MR+NS

Dividing both sides by NQ, we get:

NQRS​=NQMR​+NQNS​

Since PQ = NQ + MR, we can substitute:

PQ−MRRS​=1 (because MR/NQ and NS/NQ come from the ratios relating segments on the bases to those on the diagonals)

Substituting RS with its value (28) and rearranging the equation:

PQ = MR + PQ−MR28​

Simplifying and Solving: This equation might seem complex at first glance, but it can be simplified. We can multiply both sides by PQ - MR to get:

PQ^2 - MRP = 28

We can rewrite MRP as (PQ - RS) * RS using the expression for RS from step 6. This gives us:

PQ^2 - (PQ - 28) * 28 = 28

Expanding the equation and rearranging:

PQ^2 - 28PQ + 784 = 28

PQ^2 - 28PQ - 756 = 0

Factoring the equation:

(PQ - 42)(PQ + 18) = 0

Therefore, the possible values for PQ are 42 and -18. Since PQ represents the length of a base of a trapezoid, it cannot be negative. So, the only valid solution for PQ is:

PQ = 42

Let's find the sum of the squares of the roots of the expression: 18x2+21x−400

We can find the sum of the squares of the roots of a quadratic expression by factoring the expression and using the fact that the sum of the squares of the roots is equal to the coefficient of the quadratic term (the x2 term) plus twice the product of the linear term's coefficient (the x term) and the constant term.

Steps to solve: 1. Factor the expression: 18x2+21x−400

We can factor the expression using the sum-product pattern and extracting common factors.

2. Find the sum of the squares of the roots: Once we have the factored expression, we can identify the coefficient of the quadratic term and the product of the linear term's coefficient and the constant term.

Solution: Following the steps above, we get the factored expression: (6x−25)(3x+16)

The coefficient of the quadratic term is 18 and the product of the linear term's coefficient and the constant term is (−21)(−400)=8400.

Therefore, the sum of the squares of the roots is: 18+2(8400)=16818

Answer: The sum of the squares of the roots of the expression is 16818.

There are two main cases to consider when distributing the candy:

Case 1: The twins get the same amount of candy (more than 0 and less than all the candy)

Distribute identical candy to the twins: We can choose how many candies each twin gets in 11 ways (from 1 candy each to 6 candies each).

Distribute the remaining candy to the other 3 children: There are 13 - (2 * number of candies for twins) candies remaining.

We can distribute these candies among the 3 children in ways, following the stars and bars method (arranging 3 identical stars and (number of candies remaining - 1) non-identical bars).

Case 2: The twins each get 0 candies

Distribute all 13 candies to the other 3 children: We can distribute these candies in ways, following the stars and bars method (arranging 3 identical stars and (number of candies remaining - 1) non-identical bars).

Adding the number of ways for both cases gives us the total number of distributions.

Calculating the total ways

Case 1 ways:

Number of candy choices for twins (11 ways)

Ways to distribute remaining candy (dependent on the number chosen for twins)

Case 2 ways:

Ways to distribute candy to 3 children (without twins getting any)

For Case 1, we need to sum the ways to distribute the remaining candy for each candy choice for the twins (from 1 to 6). However, calculating this for each case can be cumbersome.

Simplifying Case 1 calculations

Notice that the ways to distribute the remaining candy only depends on the total number of candies remaining after giving the twins their share (and not on how many candies each twin got).

So, we can calculate the ways to distribute the candy for each total number of remaining candies (from 7 to 1) and reuse those values for all candy choices in Case 1 that result in the same total number of remaining candies.

Calculating reusable distribution ways

There are 3 ways to distribute 1 candy (1 child gets it, 2 children get it, or all 3 get 1 each).

There are 6 ways to distribute 2 candies (1 child gets 2, another gets none, or 2 children get 1 each).

There are 10 ways to distribute 3 candies (1 child gets 3, another gets none, 2 children get 1 each, or all 3 get 1 each).

We can continue using the stars and bars method for higher numbers of candies.

Total number of distributions

Case 1 ways:

We reuse the calculated ways to distribute remaining candy based on the total number remaining (from 7 ways for 1 candy remaining to 10 ways for 3 candies remaining).

This gives us a total of (3 + 6 + 10) * 11 = 198 ways (11 ways for choosing candy for twins multiplied by the sum of ways to distribute remaining candy (from 1 to 3 candies remaining for the twins).

Case 2 ways:

We can distribute candy to 3 children in 455 ways (using stars and bars for 13 candies and 2 dividers).

Total ways = Case 1 ways + Case 2 ways = 198 ways + 455 ways = 653 ways

Therefore, there are 653 ways to distribute 13 identical pieces of candy to 5 children if the two youngest (who are twins) receive the same amount of candy (and each can get between 0 and 6 candies).