## Tangent Lines

Suppose to co-planar circles $C_1$ and $C_2$ have no points in common. Determine how many lines, if any, are tangent to both $C_1$ and $C_2$?
Source: NCTM Mathematics Teacher, October 2006

Solution

When one circle is inside the other, there are no tangent lines to both circles. When the two circles are side by side, there are exactly four tangent lines to both circles.

Answer: $0$ or exactly $4$

## Wins and Losses

A person starting with $\256$ makes eight bets and wins exactly four times. The wins and losses occur in random order. If each wager is for half the money she has at the time of the bet, how much money will she win or lose?
Source: NCTM Mathematics Teacher, October 2006

Solution
Suppose the eight bets result in the following arbitrary sequence of wins and losses
$1\quad 256/2=128\quad\:\:\textrm{win}\quad 256+128=384$
$2\quad 384/2=192\quad\:\:\textrm{loss}\quad 384-192=192$
$3\quad 192/2=96\quad\:\:\:\:\textrm{win}\quad 192+96=288$
$4\quad 288/2=144\quad\:\:\textrm{loss}\quad 288-144=144$
$5\quad 144/2=72\quad\:\:\:\:\textrm{win}\quad 144+72=216$
$6\quad 216/2=108\quad\:\:\textrm{loss}\quad 216-108=108$
$7\quad 108/2=54\quad\:\:\:\:\textrm{loss}\quad 108-54=54$
$8\quad 54/2=27\quad\:\:\:\:\:\:\textrm{win}\quad 54+27=81$
The gambler loses $256-81=\175$.
What if we loaded all the four wins consecutively in the beginning so that she wins big and loses little money?
$1\quad 256/2=128\quad\:\:\textrm{win}\quad 256+128=384$
$2\quad 384/2=192\quad\:\:\textrm{win}\quad 384+192=576$
$3\quad 576/2=288\quad\:\:\textrm{win}\quad 576+288=864$
$4\quad 864/2=432\quad\:\:\textrm{win}\quad 864+432=1296$
$5\quad 1296/2=648\quad\textrm{loss}\quad 1296-648=648$
$6\quad 648/2=324\quad\:\:\textrm{loss}\quad 648-324=324$
$7\quad 324/2=162\quad\:\:\textrm{loss}\quad 324-162=162$
$8\quad 162/2=81\quad\:\:\:\:\textrm{loss}\quad 162-81=81$
Alas! The wins are big and so are the losses. She still loses $\175$, but why lose the same amount in the two different scenarios?
Let $x$ = money available before a bet. After a win she will have more money, $x+x/2=x(3/2)$, that is $3/2$ as much as before; after a loss she will have less money $x-x/2=x(1/2)$, that is $1/2$ as much. We track the ups and downs of money using this multiplicative rule as follows
$1\quad x/2\qquad\quad\:\:\,\textrm{win}\:\:\:\:\: x(3/2)$
$2\quad x(3/4)\quad\:\:\:\:\:\,\textrm{loss}\quad\, x(3/2)(1/2)$
$3\quad x(3/8)\quad\:\:\:\:\:\,\textrm{win}\quad\, x(3/2)^2(1/2)$
$4\quad x(9/16)\quad\:\:\:\:\textrm{loss}\quad x(3/2)^2(1/2)^2$
$5\quad x(9/32)\quad\:\:\:\:\textrm{win}\quad x(3/2)^3(1/2)^2$
$6\quad x(27/64)\quad\:\:\textrm{loss}\quad x(3/2)^3(1/2)^3$
$7\quad x(27/128)\quad\textrm{loss}\quad x(3/2)^3(1/2)^4$
$8\quad x(27/256)\quad\textrm{win}\quad x(3/2)^4(1/2)^4$
She starts out with $x=256$ and ends up with
$x(3/2)^4(1/2)^4=x(81/256)$
$=256(81/256)$
$=\81$
Though the wins and losses occur in random order, she always loses $256-81=\175$.

Answer: Will lose $\175$

## Nested Square Roots 2

Find the value of $\sqrt{16+\sqrt{16+\sqrt{16+\cdots}}}$
Source: NCTM Mathematics Teacher, October 2006

Solution
Let $x=\sqrt{16+\sqrt{16+\sqrt{16+\cdots}}}$
By substitution
$x=\sqrt{16+x}$
Squaring both sides
$x^2=16+x$
$x^2-x-16=0$
Using the quadratic formula to solve for $x$
$x=\dfrac{1\pm\sqrt{65}}{2}$
$\dfrac{1-\sqrt{65}}{2}$ is an extraneous solution because the expression cannot be negative.
$\sqrt{16+\sqrt{16+\sqrt{16+\cdots}}}=\dfrac{1+\sqrt{65}}{2}$

Answer: $\dfrac{1+\sqrt{65}}{2}$

## Digital Clock

The display on a digital clock reads $6\!:\!38$. What will the clock display twenty-seven digit changes later?
Source: NCTM Mathematics Teacher, October 2006

Solution

The figure shows there are four digit changes from $6\!:\!38$ to $6\!:\!42$ corresponding to four minute intervals. Thus, $27$ changes later $27$ minutes will have passed. The clock will display $6\!:\!38+0\!:\!27=7\!:\!05$.

Answer: $7\!:\!05$

## Waiting in Line

In a movie theater line, $x$ people are behind Mark, who is $y$ places in front of Sam. If there are $z$ people in front of Sam, how many people are in line?
Source: NCTM Mathematics Teacher, October 2006

Solution

The figure illustrates the specifics of the line. The dots represent the people waiting in line starting from the left and Mark and Sam are randomly placed in the line.
$x$ = number of people behind Mark
$y$ = Mark is $y$ places in front of Sam
$z$ = number of people in front of Sam
Let $u$ represent Sam and the number of people behind him.
Number of people in line = $z+u$
$u=x-y+1$
$z+u=z+x-y+1$

Answer: $z+x-y+1$

Alternative solution
$z$ = number of people in front of Sam excluding Sam
$x-y$ = number of people behind Sam excluding Sam
Total number of people waiting in line = $z+x-y+1$.

## New Operation

If $L$@$K=L+K/L$, then find $L$@$(L$@$K)$.
Source: NCTM Mathematics Teacher, October 2006

Solution
$L$@$(L$@$K)=L$@$(L+K/L)$
$=L+(L + K/L)/L$
$=L+L/L + K/L^2$
$=L+1+K/L^2$

Answer: $L+1+K/L^2$

## TV Game Show

On a television show, a player receives $5$ points for answering an easy question and $11$ points for a hard one. What is the largest integer that cannot be a contestant’s total score of the game?
Source: NCTM Mathematics Teacher, October 2006

Solution
Let $e$ represent the number of easy questions and $h$ the number of hard questions. We want to know if some integers cannot be written as $5e+11h$?
Let’s experiment with a random integer like $69$. If we divide $69$ by $11$ we obtain $69=11(6)+3$ which makes us think that $69$ cannot be written as $5e+11h$. But we could also subtract $11$ repeatedly
$69-11=58$
$58-11=47$
$47-11=36$
$36-11=25$
and discover that $69=5(5)+11(4)$.
We get the same results by subtracting $5$ instead of $11$
$69-5=64$
$64-5=59$
$59-5=54$
$54-5=49$
$49-5=44$
$69=5(5)+11(4)$
Some integers that cannot be a score are $29,34,39$
$29-11=18$
$18-11=7$
$34-11=23$
$23-11=12$
$12-11=1$
$39-11=28$
$28-11=17$
$17-11=6$
Is $39$ the largest integer that cannot be a score?
$40=5(8)+11(0)$
$41=5(6)+11(1)$
$42=5(4)+11(2)$
$43=5(2)+11(3)$
$44=5(0)+11(4)$
$45=5(9)+11(0)$
$46=5(7)+11(1)$
$47=5(5)+11(2)$
$48=5(3)+11(3)$
$49=5(1)+11(4)$
What about $50\,\textrm{-}\,59$?
$50=40+10=5(10)+11(0)$
$51=41+10=5(8)+11(1)$
$52=42+10=5(6)+11(2)$
$53=43+10=5(4)+11(3)$
$54=44+10=5(2)+11(4)$
$55=45+10=5(11)+11(0)$
$56=46+10=5(9)+11(1)$
$57=47+10=5(7)+11(2)$
$58=48+10=5(5)+11(3)$
$59=49+10=5(3)+11(4)$
The same pattern follows for larger integers
$60=50+10,70=60+10$, etc.
Thus, $39$ is the largest integer that cannot be a score.

Answer: $39$

## 50 Percent Larger

If $a$ is $50$ percent larger than $c$ and $b$ is $25$ percent larger than $c$, then $a$ is what percent larger than $b$?
Source: NCTM Mathematics Teacher, October 2006

Solution

For ease in argument we arbitrarily draw $c$ with $4$ squares. Then, $a$ must have $6$ squares since it is $50\%$ larger than $c$ and $b$ must have $5$ squares since it is $25\%$ larger than $c$.

How much is $a$ larger than $b$ in percent? The figure clearly shows that $a$ is larger than $b$ by $1$ square which is $1/5$ of $b$ or $20\%$.

Answer: $20\%$

Alternative solution 1
$a=c+c/2=1.5c$
$b=c+c/4=1.25c$
$\dfrac{a-b}{b}=\dfrac{1.5c-1.25c}{1.25c}=.20$ or $20\%$

Alternative solution 2
$a=1.5c$
$b=1.25c$ which implies that $c=b/1.25$
Calculate $a$ in terms of $b$
$a=1.5(b/1.25)$
$=1.2b$
$=(1+.20)b$

## A Crow Flies

A crow flies (always in straight line) from its nest $10$ miles north and $7$ miles east of the nest. At that point, it sees a scarecrow, and so it turns and flies to a point $4$ miles farther north and $5$ miles east. From there it flies directly back to the nest. To the nearest mile, what is the total distance the crow flies?
Source: NCTM Mathematics Teacher, October 2006

Solution

$AB^2=10^2+7^2=149$
$BC^2=4^2+5^2=41$
$AC^2=14^2+12^2=340$
$AB+BC+AC=\sqrt{149}+\sqrt{41}+\sqrt{340}=37.05$ miles
To the nearest mile, the crow flies a total distance of $37$ miles.

Answer: $37$ miles

If $(x/2-x/6)$ is an  integer, show why $x$ must be a multiple of $3$.
$x/2-x/6=x/3$ which is a rational number. Clearly, for $x/2-x/6$ to be an integer $x$ must be a multiple of $3$.