## 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) $x-|y|$

## Ball and Cube

Suppose that the volume of a ball is equal to $\pi/6$ times the volume of a cube. What is the ratio of the surface area of the ball to that of the cube?
Source: NCTM Mathematics Teacher, September 2006

Solution
Let $V_b$ represent the volume of the ball and $V_c$ represent the volume of the cube
$V_b=(\pi/6)V_c$
Let $r$ represent the radius of the ball and $s$ represent the length of the side of the cube
$V_b=(4/3)\pi r^3$ and $V_c=s^3$
$(4/3)\pi r^3=(\pi/6)s^3$
Multiply both sides by $6$
$8\pi r^3=\pi s^3$
Divide both sides by $\pi$
$8r^3=s^3$
$\sqrt[3]{8r^3}=\sqrt[3]{s^3}$
$2r=s$
Let $S_b$ represent the surface area of the ball and $S_c$ represent the surface area of the cube
$S_b=4\pi r^2$ and $S_c=6s^2=6(2r)^2=24r^2$
Thus, $S_b$ is to $S_c$ as $4\pi r^2$ is to $24r^2$ or $\pi$ is to $6$.

Answer: $\pi\!:\!6$

## Which Triangle is Largest?

Of the following triangles given by the lengths of their sides, which one has the greatest area?
(a) $5,12,12$
(b) $5,12,13$
(c) $5,12,14$
(d) $5,12,15$
(e) $5,12,16$
Source: NCTM Mathematics Teacher, September 2006

Solution
Heron’s Area Formula: area = $\sqrt{s(s-x)(s-y)(s-z)}$ where $s=(1/2)(x+y+z)$ and $x,y,z$ are the lengths of the sides.
(a) $s=(1/2)(5+12+12)=14.5$
area = $\sqrt{14.5(14.5-5)(14.5-12)(14.5-12)}=29.34$
(b) $s=(1/2)(5+12+13)=15$
area = $\sqrt{15(15-5)(15-12)(15-13)}=30$
(c) $s=(1/2)(5+12+14)=15.5$
area = $\sqrt{15.5(15.5-5)(15.5-12)(15.5-14)}=29.23$
(d) $s=(1/2)(5+12+15)=16$
area = $\sqrt{16(16-5)(16-12)(16-15)}=26.53$
(e) $s=(1/2)(5+12+16)=16.5$
area = $\sqrt{16.5(16.5-5)(16.5-12)(16.5-16)}=20.66$
Triangle (b) $5,12,13$ has the largest area equal to $30$ square units.

Answer: (b) $5,12,13$

Alternate solution

Since $5^2+12^2=13^2$, triangle $DBC$ is a right triangle. All other triangles are either acute or obtuse. Area of triangle $DBC=(1/2)\;5\times 12\times\mathrm{sin}\;90^\circ=30$ is the largest area.

## Area of a Disc

If the area of a disc inscribed in a square is $36\pi$ square centimeters, what is the area of the square?
Source: NCTM Mathematics Teacher, September 2006

Solution

Area of disc = $\pi r^2$
$36\pi=\pi r^2$
$36=r^2$
$r=6$
Area of square = $(2r)^2=12^2=144$

Answer: $144$ square centimeters

## Product of Consecutive Integers

How many integer pairs $(m,n)$ satisfy the equation $m(m+1)=2^n$?
Source: NCTM Mathematics Teacher, September 2006

Solution
We construct products of positive consecutive integers $m(m+1)$ and see which ones if any can be expressed as a power of $2$.
$1(2)=2=2^1$
$2(3)=6=2\cdot 3$
$3(4)=12=2^2\cdot 3$
$4(5)=20=2^2\cdot 5$
$5(6)=30=2\cdot 3\cdot 5$
$6(7)=42=2\cdot 3\cdot 7$
$7(8)=56=2^3\cdot 7$
$8(9)=72=2^3\cdot 3^2$
$9(10)=90=2\cdot 3^2\cdot 5$
$10(11)=2\cdot 5\cdot 11$
$\cdots$
$1(2)$ is the only such product and there is no other product because as the integers grow bigger and bigger, more and more prime numbers are introduced into the products making $m(m+1)=2^n$ impossible.
$1(2)=1(1+1)=2^1$ yields the first solution $(m,n)=(1,1)$ and $-2(-1)=-2(-2+1)=2^1$ yields the second solution $(m,n)=(-2,1)$.

Answer: $2$

Alternative solution 1
We construct successive positive powers of $2$ and see which ones if any can be factored into a pair of factors that are different by $1$.
$2^0=1=1(1)$
$2^1=2=1(2)$
$2^3=8=2(4)$
$2^4=16=2(8)=4(4)$
$\cdots$
$2^1=1(2)$ is the only such power.

Alternative solution 2
$m(m+1)=2^n$
$m^2+m-2^n=0$
Solving the quadratic equation by completing the square for $n=1$
$m^2+m-2^1=0$
$m^2+2m-m-2=0$
$m(m+2)-1(m+2)=0$
$(m+2)(m-1)=0$
$m=-2$ and $m=1$

Alternative solution 3

The figure shows the graphs of $y=x(x+1)$ and of $y=2^x$. They intersect at two points one of which has integer coordinates $(1,2)$ leading to $1(1+1)=2^1$.

## Real Solutions (x,y)

How  many real solutions $(x,y)$ are there that satisfy the two equations $x^2+y^2=30$ and $4y^2-x^2=100$?
Source: NCTM Mathematics Teacher, September 2006

Solution

The graph of $x^2+y^2=30$ is a circle centered at the origin and radius = $\sqrt{30}$. The graph of $4y^2-x^2=100$ is a hyperbola with vertices at $(0,5)$ and $(0,-5)$. The two graphs intersect at four points. There are four real solutions $(x,y)$ that satisfy the two equations.

Answer: $4$

Alternate solution
$x^2+y^2=30$
$4y^2-x^2=100$
————————
$5y^2=130$
$y^2=26$
$y=\pm\sqrt{26}$
Substitute the value of $y^2$ into the first equation
$x^2+26=30$
$x^2=4$
$x=\pm 2$

## Twin Liars

Kerry and Kelly are twins. One of them lies on Mondays, Tuesdays, and Wednesdays and tells the truth on the other days of the week. The other lies on Thursdays, Fridays, and Saturdays and tells the truth on the other days of the week. On which day of the week did they have this conversation?
Kerry: I lie on Saturdays.
Kelly: I will lie tomorrow.
Kerry: I lie on Sundays.
Source: NCTM Mathematics Teacher, September 2006

Solution
There are two types of liars, the Mon-Tue-Wed liar and the Thu-Fri-Sat liar. We need to determine who is who? Clearly, Kerry is lying when he said that “I lie on Sundays” since both of them tell the truth on Sundays. Therefore, his statement “I lie on Saturdays” is false; he tells the truth on Saturdays and so Kerry is the Mon-Tue-Wed liar and Kelly is the Thu-Fri-Sat liar.

Since we know Kerry is lying, the conversation can only take place on Monday, Tuesday, or Wednesday.

Suppose it was Monday. Since Kelly is telling the truth on Mondays, the statement “I will lie tomorrow” (Tuesday) is true, which characterizes her as a Tue. liar. But that is not possible, because she is a Thu-Fri-Sat liar.

Suppose it was Tuesday. Since Kelly is telling the truth on Tuesdays, the statement “I will lie tomorrow” (Wednesday) is true, which characterizes her as a Wed. liar. But that is not possible, because she is a Thu-Fri-Sat liar.

Suppose it was Wednesday. Since Kelly is telling the truth on Wednesdays, the statement “I will lie tomorrow” (Thursday) is true, which characterizes her as a Thu. liar. There is no contradiction here, she is indeed a Thu-Fri-Sat liar.