## Shift Left

Given the equation $x^3-2x^2+x-3=0$, find an equation whose roots are each $2$ less than the roots of the given equation.
Source: NCTM Mathematics Teacher, August 2006

Solution

If we graph the given equation as a function $y=x^3-2x^2+x-3$, we see that it has one real root at $x\approx 2.1746$. The other two roots are complex and not shown. To get a new equation that has also three roots each $2$ less than the roots of the given equation, all we have to do is shift the graph to the left by $2$ units. Algebraically, it means replacing $x$ by $x-(-2)=x+2$
$y=(x+2)^3-2(x+2)^2+(x+2)-3$
$=x^3+4x^2+5x-1$

The new equation is $x^3+4x^2+5x-1=0$

Answer: $x^3+4x^2+5x-1$

## Tangent Circles

Consider the circles that have radii $4\sqrt{5}$ and are tangent to the line $x-2y=20$ at the point $(6,-7)$. Find the sum of the $x$ coordinates of the centers of the circles.
Source: NCTM Mathematics Teacher, August 2006

Solution

For the two circles to be tangent at point $(6,-7)$ located on line $l_1$, line $l_2$ passing through the centers must be perpendicular to $l_1$. Given that the equation of $l_1$ is $x-2y=20$, the equation of $l_2$ is found to be $y=-2x+5$.
Let $(h,k)$ represent the coordinates of the center of one of the circles.
$(x-h)^2+(y-k)^2=(4\sqrt{5})^2$    — equation of the circle
$(6-h)^2+(-7-k)^2=80$          — point $(6,-7)$ is on the circle
$k=-2h+5$                                  — center $(h,k)$ is on $l_2$
Substitute the value of $k$ into the quadratic equation
$(6-h)^2+(-7-(-2h+5))^2=80$
Expand and simplify
$(6-h)^2+(-7+2h-5)^2=80$
$(6-h)^2+(-12+2h)^2=80$
$36-12h+h^2+144-48h+4h^2=80$
$5h^2-60h+100=0$
Divide both sides by $5$
$h^2-12h+20=0$
If $a$ and $b$ are the zeros of a quadratic equation, $(x-a)(x-b)=0$ leads to $x^2-(a+b)x+ab=0$. The sum of the $h$ solutions = $12$ and their product = $20$. The sum of the $x$-coordinates of the circle = $12$.

Answer: $12$

## Father and Son Eat Together

A father and son eat meals together. If the son eats twice as fast as the father and the father eats a meal in $45$ minutes, how many meals can the son eat in $3$ hours?
Source: NCTM Mathematics Teacher, August 2006

Solution

The diagram displays the $3$-hour timeline in $15$-minute intervals. If the father eats a meal in $45$ minutes, he can finish $4$ meals in $3$ hours. If the son eats twice as fast as the father, he can eat $8$ meals in the same amount of time.

Answer: $8$

## Round Table

A round table can be made square by dropping the four leaves. If a side of a square table measures $36$ inches, to the nearest hundredth of a square inch, how much smaller is the area of the table when the leaves are down than when the leaves are up?
Source: NCTM Mathematics Teacher, August 2006

Solution

The radius $r=18\sqrt{2}$ because the triangle is a $45^\circ\!-\!45^\circ\!-\!90^\circ$ triangle.
Area of the circle =$\pi(18\sqrt{2})^2$
Area of square =$36^2$
Difference in areas =$\pi(18\sqrt{2})^2-36^2\approx 739.75$ square inches.

Answer: $739.75$ square inches

## Largest Equilateral Triangle in a Square

How long is the side of the largest equilateral triangle that can be inscribed in a square with side length $1$?
Source: NCTM Mathematics Teacher, August 2006

Solution

A first guess is an equilateral triangle of side length $1$. Could there be a larger triangle?

The hypotenuse in the right triangle certainly has length greater than $1$. But how much greater can it get and still be able to form an equilateral triangle?

Let $s$ be the side length of the equilateral triangle. By the Pythagorean theorem we have the following equations
$s^2=(1-x)^2+(1-x)^2\qquad (1)$
$s^2=1^2+x^2\qquad\qquad\qquad\;\;\;\; (2)$
Compare the two equations
$2(1-x)^2=1+x^2$
$2(1-2x+x^2)=1+x^2$
$2-4x+2x^2=1+x^2$
$x^2-4x+1=0$
Solve for $x$ by completing the square
$x^2-4x+4=3$
$(x-2)^2=3$
$x-2=\pm\sqrt{3}$
We reject $x=2+\sqrt{3}$ because $1-x<0$.
Thus, $x=2-\sqrt{3}$
Substitute the value of $x$ into Eq. $(2)$
$s^2=1+(2-\sqrt{3})^2$
$=1+(4-4\sqrt{3}+3)$
$=8-4\sqrt{3}$
$=4(2-\sqrt{3})$
$s=\pm 2(\sqrt{2-\sqrt{3}})$
$2(\sqrt{2-\sqrt{3}})\approx 1.0353$ not significantly greater than $1$.

Answer: $2(\sqrt{2-\sqrt{3}})$

## Third Root of Equation

Two of the roots of the equation $2x^3-3x^2+px+q=0$ are $3$ and $-2$. Find the third root of the equation.
Source: NCTM Mathematics Teacher, August 2006

Solution
Let $a$ be the third root of the equation. We write the equation in factored form
$2x^3-3x^2+px+q=(2x-6)(x+2)(x-a)$
Multiply the right-hand side
$=(2x^2+4x-6x-12)(x-a)$
$=(2x^2-2x-12)(x-a)$
$=2x^3-2x^2-12x-2ax^2+2ax+12a$
$=2x^3-(2+2a)x^2+(2a-12)x+12a$
Compare the coefficients of $x^2$
$3=2+2a$
$a=1/2$

Answer: $1/2$

Alternative solution 1
Consider a general polynomial of third degree with roots $a,b,c$ and coefficients $a_0,a_1,a_2,a_3$
$a_3x^3+a_2x^2+a_1x+a_0=(x-a)(x-b)(x-c)$
Multiplying out the product on the right-hand side yields
$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$
$a+b+c=a_2$ provided that $a_3=1$
Divide $2x^3-3x^2+px+q$ by $2$
$x^3-(3/2)x^2+(p/2)x+q/2$
Third root = $3/2-3-(-2)=1/2$

Alternative solution 2
Substitute the value of $x=3$ into the equation
$2(3)^3-3(3)^2+p(3)+q=0$
$27+3p+q=0$
$3p+q=-27\qquad (1)$
Substitute the value of $x=-2$ into the equation
$2(-2)^3-3(-2)^2+p(-2)+q=0$
$-28-2p+q=0$
$-2p+q=28\qquad (2)$
Multiply Eq. $(1)$ by $-1$ and add to Eq. $(2)$
$-3p-q=27$
$-2p+q=28$
——————–
$-5p=55$
$p=-11$
Substitute the value of $p$ into Eq. $(1)$
$3(-11)+q=-27$
$q=6$
We now have the cubic equation $2x^3-3x^2-11x+6=0$ with two known roots $3,-2$. We can find the third root either by long division or by synthetic division: $2x^3-3x^2-11x+6=(x-3)(x+2)(2x-1)$.
We can also graph the equation to find the third root

## Partial Fractions

When $\dfrac{19x-8}{2x^2-x-21}$ is decomposed into partial fractions (fractions that sum to the given fraction), what is the sum of the numerators when each fraction is reduced to lowest terms?
Source: NCTM Mathematics Teacher, August 2006

Solution
$\dfrac{19x-8}{2x^2-x-21}=\dfrac{19x-8}{(2x-7)(x+3)}$
$=\dfrac{A}{2x-7}+\dfrac{B}{x+3}$
$=\dfrac{A(x+3)+B(2x-7)}{(2x-7)(x+3)}$
Solving for $A,B$
$Ax+3A+2Bx-7B=19x-8$
Collecting like terms
$(A+2B)x+3A-7B=19x-8$
$A+2B=19\qquad\;\;\; (1)$
$3A-7B=-8\qquad (2)$
Multiplying Eq. $(1)$ by $7$ and Eq. $(2)$ by $2$
$7A+14B=133$
$6A-14B=-16$
————————
$13A=117$
$A=9$
Substitute the value of $A$ into Eq. $(1)$
$9+2B=19$
$B=5$
$\dfrac{19x-8}{2x^2-x-21}=\dfrac{9}{2x-7}+\dfrac{5}{x+3}$
$9+5=14$

Answer: $14$

## Equilateral Triangle and Square

A pentagon made up of equilateral triangle $ABC$ with side length $2$ on top of square $BCDE$ is inside a circle passing through points $A,D$, and $E$. Find the radius of the circle.

Source: NCTM Mathematics Teacher, August 2006

Solution

Let $F$ be the center of the circle with radius $r$ passing through points $A,D$, and $E$. Draw $\overline{AM}$ the perpendicular bisector of side $\overline{BC}$ and $\overline{AN}$ the perpendicular bisector of side $\overline{ED}$. The bisectors pass through center $F$ because it is equidistant from the endpoints $E,D$.
We have the following equations
$FN^2+1^2=r^2$     ($FND$ is a right triangle)
$FM+FN=2$     ($MCDN$ is a rectangle)
$AM+FM=r$
$\sqrt{3}+FM=r$     ($AMC$ is a $30^\circ\!-\!60^\circ\!-\!90^\circ$ triangle)
For ease in notation, we rewrite the equations with  $FM=x,FN=y$
$y^2+1=r^2\qquad (1)$
$x+y=2\qquad\;\;\; (2)$
$\sqrt{3}+x=r\qquad (3)$
Squaring both sides of Eq. $(3)$
$(\sqrt{3}+x)^2=r^2$
$3+2\sqrt{3}x+x^2=r^2$
Substitute the value of $r^2$ from Eq. $(1)$
$3+2\sqrt{3}x+x^2=y^2+1$
Substitute the value of $y$ from Eq. $(2)$
$3+2\sqrt{3}x+x^2=(2-x)^2+1$
$3+2\sqrt{3}x+x^2=4-4x+x^2+1$
Simplifying and collecting like terms
$2\sqrt{3}x+4x=2$
$(\sqrt{3}+2)x=1$
$x=1/(\sqrt{3}+2)=(\sqrt{3}-2)/(3-4)=2-\sqrt{3}$
Substitute the value of $x$ into Eq. $(3)$
$\sqrt{3}+(2-\sqrt{3})=r$
$r=2$
The radius of the circle equals $2$.

Answer: $2$

Alternative solution

Draw $\overline{AM}$ the perpendicular bisector of $\overline{BC}$. It passes through center $F$ because $F$ is equidistant from the endpoints $E$ and $D$. Since $B$ is equidistant from the endpoints $A$ and $E$, $B$ is on the perpendicular bisector of $\overline{AE}$. Similarly, since $F$ is equidistant from the endpoints $A$ and $E$, $F$ is on the perpendicular bisector of $\overline{AE}$.
In an isosceles triangle the perpendicular bisector is also the angle bisector. Hence the measure of $\angle{ABF}=(90^\circ\!+60^\circ)/2=75^\circ$ which implies that the measure of $\angle{MBF}=75^\circ\!-60^\circ=15^\circ$. Therefore, the measure of $\angle{BFM}=90^\circ\!-15^\circ=75^\circ$. Triangle $ABM$ is an isosceles triangle and consequently $AF=AB=2$.

## Children Sharing Money

Al, Bee, Cecil, and Di have $\16,\24,\32$, and $\48$, respectively. Their father proposed that Al and Bee share their wealth equally, and then Bee and Cecil do likewise, and then Cecil and Di. Their mother’s plan is the same except that Di and Cecil begin by sharing equally, then Cecil and Bee, and then Bee and Al. Determine the number of children who end up with more money under their father’s plan than under their mother’s plan.
Source: NCTM Mathematics Teacher, August 2006

Solution
Father’s plan
$(\mathrm{Al}+\mathrm{Bee})/2=(16+24)/2=20$
$(\mathrm{Bee}+\mathrm{Cecil})/2=(20+32)/2=26$
$(\mathrm{Cecil}+\mathrm{Di})/2=(26+48)/2=37$
Al, Bee, and Cecil end up with more money.
Mother’s plan
$(\mathrm{Di}+\mathrm{Cecil})/2=(48+32)/2=40$
$(\mathrm{Cecil}+\mathrm{Bee})/2=(40+24)/2=32$
$(\mathrm{Bee}+\mathrm{Al})/2=(32+16)/2=24$
Only Al ends up with more money.
Two children end up with more money under father’s plan than under mother’s plan.

Answer: $2$

## Cottages on a Straight Road

There are four cottages on a straight road. The distance between Ted’s and Alice’s cottages is $3$ km. Both Bob’s and Carol’s cottages are twice as far from Alice’s as they are from Ted’s. Find the distance between Bob’s and Carol’s cottages in kilometers.
Source: NCTM Mathematics Teacher, August 2006

Solution

Suppose the four cottages of Bob, Ted, Carol, and Alice are located on a straight line with respective coordinates  $b,t,c$, and $a$ such that $0.
Recall that $|x-y|=$ distance between real numbers $x$ and $y$ on the number line.

$|t-a|=$ distance between Ted’s and Alice’s cottages
$|b-a|=$ distance between Bob’s and Alice’
s cottages
$|c-a|=$ distance between Carol’s and Alice’s cottages
$|c-t|=$ distance between Carol’s and Ted’s cottages
Given that $|t-a|=3,|b-a|=2|b-t|$, and $|c-a|=2|c-t|$, we want to find $|b-c|$ the distance between Bob’s and Carol’s cottages.
$|t-a|=3$ implies $a-t=3\qquad\qquad\:\: (1)$
Calculate $|b-t|$ the distance between Bob’s and Ted’s cottages
$|b-a|=2|b-t|$ implies $a-b=2(t-b)$
Add $(b-t)$ to both sides of the equation
$a-b+(b-t)=2t-2b+(b-t)$
$a-t=t-b$
Substitute the value of $(a-t)$ from Eq. $(1)$
$3=t-b$ which implies $|b-t|=3\qquad (2)$
Calculate $|t-c|$ the distance between Ted’s and Carol’s cottages
$|t-c|+|c-a|=3$
$|t-c|+2|c-t|=3$
$3|t-c|=3$
$|t-c|=1\qquad\qquad (3)$
Calculate $|b-c|$ the distance between Bob’s and Carol’s cottages. From Eqs. $(2)$ and $(3)$
$|b-c|=|b-t|+|t-c|=3+1=4$ km

Answer: $4$ km