## Distinct Triangles

How many distinct triangles can be constructed by choosing three vertices from among the corners of a unit cube?
Source: NCTM Mathematics Teacher, February 2008

Solution

We construct four triangles from three corners of a face for a total of $6\times 4=24$ triangles.

In the plane formed by the the two diagonals $A$ and $B$ and the two sides, we construct four triangles. Similarly, the diagonals C and D produce four more triangles. Thus, the Top/Bottom, Front/Back and Left/Right faces make total of $8+8+8=24$ triangles.

Take corner $E$ as the top and form a tetrahedron with the three diagonals as the base. The tetrahedron produces four triangular faces – three are repeat, one is distinct. Hence, eight corners make $8$ distinct triangles.
The number of distinct triangles constructed by choosing three vertices from among the corners of a unit cube = $24+24+8=56$.

Answer: $56$

Alternative solution
We construct a triangle from any three corners of the cube because they are not collinear. The number of ways to choose three corners from a set of eight = $\dbinom{8}{3}=56$.

## Measure of Angle Z

Find the measure of the angle $Z$ in the diagram here, given that the two horizontal segments are parallel.

Source: NCTM Mathematics Teacher, February 2008

Solution

Complete a seven-sided concave polygon by drawing a perpendicular line to the two parallel lines as shown in the diagram. Note that the interior of the polygon is made up of five triangles. Hence the sum of the polygon’s seven interior angles = $5\times 180^\circ\!=900^\circ$. The measure of the interior angle $Z^\prime=900^\circ\!-145^\circ\!-135^\circ\!-4(90^\circ\!)=260^\circ$. So, the measure of the exterior angle $Z=360^\circ\!-260^\circ\!=100^\circ$.

Answer: $100^\circ$

Alternative solution

Draw lines parallel to the two base lines through the vertex of each angle. Apply a series of theorems on angles formed by parallel lines and transversals such as Consecutive Interior Angles, Alternate Interior Angles, to show that the measure of $Z=55^\circ+45^\circ=100^\circ$.

## Square of 111,111,111

Find the exact value of $(111,111,111)^2$.
Source: NCTM Mathematics Teacher, February 2008

Solution
$11^2=121$
$111^2=12321$
$1111^2=1234321$
$11111^2=123454321$
$\cdots$
$111111111^2=12345678987654321$

Answer: $12,\!345,\!678,\!987,\!654,\!321$

Alternative solution
Multiply the numbers manually using the old-fashioned multiplication algorithm. It should be straightforward because there is no “carry”.

## Two Even Digits

The number $2008$ is a four-digit number having two zeros and two even digits that are not zeros. How many four-digit numbers besides $2008$ can also be described this way?
Source: NCTM Mathematics Teacher, February 2008

Solution
First, choose a way to place the two zeros. We have three possible ways to do that, $-0\,0-,--0\,0$, and $-0-0$ (we cannot place zero as a leading digit). Second, place the two even digits in the two empty places. We have $4\times 4=16$ possible ways to place them in light of the four even digits $2,4,6,8$. Thus, there are $3\times 16=48$ four-digit numbers that have two zeros and two even digits. If we do not count $2008$, there are 47 such numbers.

## Product of Factorials

Solve for $N$ if $6!\times 7!=N!$.
Source: NCTM Mathematics Teacher, February 2008

Solution
$6!\times 7!=(6\times 5\times 4\times 3\times 2\times 1)7!$
$=(6\times 5\times 3)8\times 7!$
$=(5\times 18)8\times 7!$
$=(5\times 2)9\times 8\times 7!$
$=10\times 9\times 8\times 7!$
$=10!$
$N=10$

Answer: $10$

## Four Congruent Rectangles

Four congruent rectangles are arranged as shown in the figure to form a large rectangle. The area of the large rectangle is $768$. What is the area of a square that has the same perimeter as the large rectangle?

Source: NCTM Mathematics Teacher, February 2008

Solution

Area of a smaller congruent rectangle = $ab$.
$768=4(ab)$
$ab=192$
Area of the large rectangle = $(a+b)a=a^2+ab$.
$768=a^2+192$
$a^2=576$
$a=24$
$b=192/a=192/24=8$
Perimeter of the large rectangle
$2(a+b)+2a = 2(24+8)+2(24)=112$
Area of square = $(112/4)^2=784$

Answer: $784$ square units

Alternative solution
The width of the large rectangle $a=3b$ and the length $a+b=3b+b=4b$.
$768=(4b)(3b)$
$768=12b^2$
$b^2=64$
$b=8$
$a=3b=3(8)=24$

## Sum of Six Angles

Find the value in degrees of the sum of the six angles indicated in the diagram.

Source: NCTM Mathematics Teacher, February 2008

Solution
The three triangles contain nine angles, three of which form a straight angle. Hence, the sum of the six angles indicated in the diagram is equal to the sum of the nine angles minus the straight angle = $3\times 180-180=360^\circ$.

Answer: $360^\circ$

## The Second Number

The sum of three numbers is $98$. The first number is $2/3$ of the second, and the second is $5/8$ of the third. What is the second number?
Source: NCTM Mathematics Teacher, February 2008

Solution
Let $a,b,c$ be the first, second, and third number. Substitute the value of $a=(2b)/3$ and $c=(8b)/5$ into the sum equation
$a+b+c=98$
$(2b)/3+b+(8b)/5=98$
$(10b+15b+24b)/15=98$
$49b=98\times 15$
$b=30$

Answer: $30$

## Prime Divisor of Factorials

Determine the largest prime divisor of $87!+88!$
Source: NCTM Mathematics Teacher, February 2008

Solution
$87!+88!=87!+88(87!)=87!(1+88)=89(87\cdot 86\cdot 85\cdots 2\cdot 1)$
$89$ is the largest prime divisor.

Answer: $89$

## Nickels’ Worth

A stack of $100$ nickels is $6.25$ inches high. To the nearest cent, how much would a stack of nickels $8$ feet high be worth?
Source: NCTM Mathematics Teacher, February 2008

Solution
Let $x$ be the worth in dollars of a stack of nickels $8$ feet high ($8\times 12$ inches). $100$ nickels are worth $5$ dollars. We have the following proportion
$\dfrac{5}{6.25}=\dfrac{x}{8\times 12}$
$x=\dfrac{5\times 8\times 12}{6.25}=\char36 76.80$

Answer: $\char36 76.80$