## Product of x-Intercepts

The line $L$ with equation $y=mx+b$ and $M$ with equation $y=bx+m$ are perpendicular. If neither passes through the origin, find the product of their $x$-intercepts.

Solution
Let point $(x_L,0)$ represent the $x$-intercept of line $L$
$0=mx_L+b$
$mx_L=-b$
$x_L=-b/m$
Let point $(x_M,0)$ represent the $x$-intercept of line $M$
$0=bx_M+m$
$bx_M=-m$
$x_M=-m/b$
Product of the $x$-intercepts
$x_Lx_M=(-b/m)(-m/b)=1$

Answer: $1$

Alternative solution
Slope of line $L=m$, slope of line $M=b$. By the property of perpendicular lines $mb=-1$.
$0=mx_L+b$
Multiply both sides by $m$
$0=m^2x_L+mb$
$0=m^2x_L-1$
$m^2x_L=1$
$x_L=1/m^2$
Similarly $x_M=1/b^2$

$x_Lx_M=1/m^2 \times 1/b^2=1/(m^2b^2)=1/(mb)^2=1/(-1)^2=1$

## Five Same Numbers

You roll five fair, six-sided dice, all at the same time. In that one roll, what is the probability of getting five of the same number?
Source: NCTM Mathematics Teacher, September 2006

Solution
When you roll five dice at the same time, $6$ possible numbers from $1$ through $6$ can come up on each of the five dice. The number of possible outcomes = $6\times 6\times 6\times 6\times 6=6^5$. There are $6$ possible ways the five dice come up with the same number: $11111,22222,33333,44444,55555,66666$. The probability of getting five of the same number in one roll = $6/6^5=1/6^4=1/1296$.

Answer: $1/1296$

Alternative solution
Suppose the event $11111$ occurs on a roll. Since the number $1$ on the first dice is independent of the number $1$ on the second dice, etc., the probability of the event $11111=1/6\times 1/6\times 1/6\times 1/6\times 1/6=1/6^5$. The probability of getting five of the same number in one roll = $1/6^5+1/6^5+1/6^5+1/6^5+1/6^5=6(1/6^5)=1/6^4=1/1296$.

## Two Equilateral Triangles

If two equilateral triangles of area $A$ intersect to form a regular hexagon, then what is the area of the hexagon?
Source: NCTM Mathematics Teacher, September 2006

Solution

If we subdivide an equilateral triangle into $9$ smaller congruent equilateral triangles as shown in the figure below, we see that the hexagon is made up of $6$ of them.

Area of a smaller triangle = $A/9$
Area of the hexagon = $6(A/9)=2A/3$

Answer: $2A/3$

## Difference of Two Squares

Which one of the following numbers cannot be expressed as the difference of the squares of two integers?
(a) $314159265$
(b) $314159266$
(c) $314159267$
(d) $314159268$
(e) $314159269$
Source: NCTM Mathematics Teacher, September 2006

Solution
Let $k$ be an odd integer and $k=m^2-n^2=(m+n)(m-n)$ for some integer $m$ and $n$. Since any odd integer is a multiple of $1$, we choose $(m-n)=1$ and write $k=(m+n)(1)=m+n$. We could have chosen $(m-n)=3,5,7,\cdots$ but that would restrict $n$ to be multiples of $3,5,7,\cdots$
Solving for $m$ and $n$ from the equations $m+n=k$ and $m-n=1$ yields
$m=(k+1)/2$
$n=(k-1)/2$
Any odd integer can be expressed as a difference of the squares of two integers. We rule out (a) $314159265$, (c) $314159267$, and (e) $314159269$. Their decomposition into a difference of the squares of two integers is shown below
(a) $314159265$
$m=(314159265+1)/2=157079633$
$n=(314159265-1)/2=157079632$
$314159265=157079633^2-157079632^2$
(c) $314159267$
$m=(314159267+1)/2=157079634$
$n=(314159267-1)/2=157079633$
$314159267=157079634^2-157079633^2$
(e) $314159269$
$m=(314159269+1)/2=157079635$
$n=(314159269-1)/2=157079634$
$314159269=157079635^2-157079634^2$

Let $k$ be an even integer and $k=m^2-n^2=(m+n)(m-n)$ for some integers $m$ and $n$. Since any even number is a multiple of $2$, we choose $(m-n)=2$ and write $k=(m+n)(2)$. Solving for $m$ and $n$ from the equations $2m+2n=k$ and $m-n=2$ yields
$m=(k+4)/4=k/4+1$
$n=(k-4)/4=k/4-1$
For an even integer $k$ to be expressed as a difference of the squares of two integers it must be divisible by $4$.
(b) $314159266$ is not divisible by $4$
(d) $314159268$ is divisible by $4$
$m=314159268/4+1=78539817+1=78539818$
$n=314159268/4-1=78539817-1=78539816$
$314159268=78539818^2-78539816^2$

Answer: (b) $314159266$

Alternate solution
Let $k$ and $k+1$ be two consecutive integers.
$(k+1)^2-k^2=k^2+2k+1-k^2=2k+1$ an odd integer.
Any odd integer can be expressed as a difference of the squares of two integers.

Let $k$ be an even integer and $k=m^2-n^2$ for some integers $m$ and $n$. For $m^2-n^2$ to be even either both $m^2$ and $n^2$ are even or both $m^2$ and $n^2$ are odd.
(1) If both $m^2$ and $n^2$ are even, then both $m$ and $n$ are even. Let $m=2i$ and $n=2j$ for some integers $i$ and $j$.
$m^2-n^2=(2i)^2-(2j)^2=4i^2-4j^2=4(i^2-j^2)$
(2) If both $m^2$ and $n^2$ are odd, then both $m$ and $n$ are odd. Let $m=2i+1$ and $n=2j+1$ for some integers $i$ and $j$.
$m^2-n^2=(m+n)(m-n)$
$=(2i+1+2j+1)(2i+1-2j-1)$
$=(2i+2j+2)(2i-2j)$
$=2(i+j+1)(2)(i-j)$
$=4(i+j+1)(i-j)$
For an even integer $k$ to be expressed as a difference of the squares of two integers, it must be a multiple of $4$.

## Two Squares

If the vertices of square $A$ are the midpoints of the edges of square $B$, then what is the ratio of the area of square $A$ to square $B$?
Source: NCTM Mathematics Teacher, September 2006

Solution

The diagonal of square $B$ divides square $A$ into two congruent rectangles, the dimensions of which are, length = $a\sqrt{2}$ and width = $(a\sqrt{2})/2$.
Area of square A = $a\sqrt{2}\times (a\sqrt{2})/2+a\sqrt{2}\times (a\sqrt{2})/2=2a^2$
Area of square B = $(2a)^2=4a^2$
Ratio of the area of square A to the area of square B is $2a^2\!:\!4a^2$ or $1\!:\!2$.

Answer: $1\!:\!2$

Alternative solution 1

Area of square A = $(a\sqrt{2})^2=2a^2$
Area of square B = $(2a)^2=4a^2$

Alternative solution 2

The dotted grid lines divide square $B$ into four little squares, each of which is further divided into two congruent isosceles right triangles. Square $B$ is made up of $8$ right triangles whereas square $A$ is made up of $4$ right triangles. Thus the ratio of the area of square $A$ to the area of square $B$ is $4\!:\!8$ or $1\!:\!2$.

Alternative solution 3

Think of the two squares as two rhombi with congruent diagonals. Area of a rhombus = $(1/2)d_1\times d_2$ where $d_1$ and $d_2$ are diagonals.
Area of square A = $(1/2)\times 2a\times 2a=2a^2$
Area of square B = $(1/2)\times 2(a\sqrt{2})\times 2(a\sqrt{2})=4a^2$

## Maximum Angle

The measures of five angles of a convex polygon are each equal to $108^\circ$. In which of the five open intervals does the maximum angle of all such polygon lie?
(a) $(105^\circ\!,\!120^\circ\!)$
(b) $(120^\circ\!,\!135^\circ\!)$
(c) $(135^\circ\!,\!150^\circ\!)$
(d) $(150^\circ\!,\!165^\circ\!)$
(e) $(165^\circ\!,\!180^\circ\!)$
Source: NCTM Mathematics Teacher, September 2006

Solution

Let $n$ represent the number of sides. Suppose $n=6$. The figure shows the convex polygon has six vertices $A,B,C,D,E,F$ and six sides. Five of the angles have an equal measure of $108^\circ$. Divide the polygon into four triangles by connecting vertex $D$ to vertices $A,B$, and $F$. By the Triangle Sum theorem, the sum of the measures of the interior angles of a triangle is $180^\circ$. The sum of the measures of all six angles of the polygon equals $4\times 180^\circ\!=720^\circ$. Hence the measure of the remaining angle $D=720^\circ\!-5(108^\circ\!)=180^\circ$.

Because angle $D$ is a straight angle, if we were to travel from vertex $C$ to vertex $E$, we would cross into the interior of the polygon contradicting the definition of a convex polygon.
Similarly for $n\ge 7$, the maximum measure of one of the remaining angles is $\ge 180^\circ$. For example, when $n=7$, the sum of the measures of the two remaining angles is $360^\circ$. Their possible measures are
Angle 1    Angle 2
$1^\circ \qquad\quad 359^\circ$
$2^\circ\qquad\quad 358^\circ$
$3^\circ\qquad\quad 357^\circ$
$\cdots\qquad\quad\cdots$
$179^\circ\qquad 181^\circ$
$180^\circ\qquad 180^\circ$
$181^\circ\qquad 179^\circ$
$\cdots\qquad\quad\cdots$
$357^\circ\qquad 3^\circ$
$358^\circ\qquad 2$
$359^\circ\qquad 1^\circ$
Thus the polygon has only five angles each measuring $108^\circ$ as the maximum value of the measures of the angles.

Answer: (a) $(105^\circ\!,\! 120^\circ\!)$

Alternative solution
By the Polygon Exterior Angles theorem, the sum of the measures of the exterior angles of a convex polygon, one angle at each vertex, is $360^\circ$. Since $5(180^\circ\!-108^\circ\!)=5(72^\circ\!)=360^\circ$, the five angles are the only angles of the convex polygon.

## Length of Hypotenuse

Consider the line $l$ containing the points $(-3,8)$ and $(6,-4)$. What is the length of the hypotenuse of the right triangle formed by the intersections of $l$ and the $x$-and $y$-axes?
Source: NCTM Mathematics Teacher, September 2006

Solution

Points $A(-3,-4),B(-3,8)$, and $C(6,-4)$ are the vertices of the right triangle $ABC$. The length of hypotenuse $\overline{BC}$ equals
$AB^2+AC^2=BC^2$
$12^2+9^2=BC^2$
$225=BC^2$
$BC=15$
The slope of $\overline{BC}$ equals
$\dfrac{8-(-4)}{-3-6}=\dfrac{12}{-9}=-\dfrac{4}{3}$
Hypotenuse $\overline{BC}$ intersects the $x$-axis at point $D(x,0)$. Using $D(x,0)$ and $C(6,-4)$ we express the slope of $\overline{DC}$ as
$\dfrac{0-(-4)}{x-6}=-\dfrac{4}{3}$
$\dfrac{4}{x-6}=-\dfrac{4}{3}$
$x-6=-3$
$x=3$
Thus, $\overline{BC}$ intersects the $x$-axis at $D(3,0)$. Observe that the four vertical lines $x=-3,x=0,x=3$, and $x=6$ intersect the $x$-axis at equal intervals. By the Proportionality theorem, if parallel lines intersect two transversals, then they divide the transversals proportionally. Given that the four vertical lines are parallel and divide the transversal $x$-axis into thirds, they also divide the transversal $\overline{BC}$ into thirds.
$BE=ED=DC=BC/3=15/3=5$

Answer: $5$ units

Alternative solution
Using points $E(0,y)$ and $B(-3,8)$ we express the slope of $\overline{EB}$ as
$\dfrac{y-8}{0-(-3)}=-\dfrac{4}{3}$
$\dfrac{y-8}{3}=-\dfrac{4}{3}$
$y-8=-4$
$y=4$
The triangle with vertices at $(0,0),D(3,0)$ and $E(0,4)$ is a $3,4,5$ right triangle.

## Ten-Card Game

A game is played with a deck of ten cards numbered from $1$ to $10$. Shuffle the deck thoroughly. Take the top card. If it is numbered $1$, you win. If it is numbered $k$, where $k>1$, then replace the card into the $k$th position from the top and draw again. You are allowed a maximum of $3$ draws before losing the game. What is the probability of winning?
Source: NCTM Mathematics Teacher, September 2006

Solution
It helps to build a deck of cards made out of cutout papers and simulate a few draws in order to understand the mechanics of the game. For example, we discover that having the card numbered $1$ in third position does not guarantee a win, unless the card numbered $2$ is not in first position.

An indirect approach to solve the problem is to figure out how many ways can we lose the game? If the card numbered $1$ is in positions $4,5,6,7,8,9$, or $10$ from the top, we will definitely lose because there is no way for it to bubble up to the top in $3$ draws. The probability for each of these seven events equals $1/10$. We will also lose if the card numbered $1$ is in third position and the card numbered $2$ is in first position. The probability for this event equals $(1/10)(1/9)$.
The probability for losing equals
$7(1/10)+(1/10)(1/9)=7/10+1/90=64/90$
Hence, the probability for winning equals
$1-(64/90)=26/90=13/45$

Answer: $13/45$

Alternative solution
A direct approach to solve the problem is to figure out how many ways can we win the game? If the card numbered $1$ is in first or second positions, we will win for sure. The probability for each of these two events equals $1/10$. If the card numbered $1$ is in third position and the card numbered $2$ is not in first position, we will also win. The probability for this event equals $(1/10)(8/9)$.
The probability of winning equals
$1/10+1/10+(1/10)(8/9)=2/10+8/90=26/90=13/45$

## Rolling Wheel

If a wheel completes exactly $5$ revolutions while rolling $80\pi$ feet, what is the diameter of the wheel?
Source: NCTN Mathematics Teacher, September 2006

Solution

The above figure shows the ground distance traveled by the rolling wheel of radius $r$ feet in $5$ revolutions.
$5(2\pi r)=80\pi$
$2r=16$ feet
The diameter of the wheel equals $16$ feet.

Answer: $16$ feet

## Smallest Quantity

If $x and $x<0$, which of the following is never greater than any of the others?
(a) $x+y$
(b) $x-y$
(c) $x+|y|$
(d) $x-|y|$
(c) $-|x+y|$
Source:NCTM Mathematics Teacher, September 2006

Solution
Suppose $x=-5$ and $y=-2$
(a) $x+y=-5+(-2)=-7$
(b) $x-y=-5-(-2)=-3$
(c) $x+|y|=-5+2=-3$
(d) $x-|y|=-5-2=-7$
(e) $-\left | x+y\right |=-\left | -7\right |=-7$
(a),(d),(e) have the smallest value $-7$

Suppose $x=-5$ and $y=2$
(a) $x+y=-5+2=-3$
(b) $x-y=-5-2=-7$
(c) $x+\left| y\right |=-5+2=-3$
(d) $x-|y|=-5-2=-7$
(e) $-\left | x+y\right |=-\left | -3\right |=-3$
(b) and (d) have the smallest value $-7$

In all cases (d) has the smallest value.

Answer: (d) $x-|y|$