2006 iTest Problems/Problem U6

The following problem is from the Ultimate Question of the 2006 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.

Problem

$x$ and $y$ are nonzero real numbers such that

$18x - 4x^2 + 2x^3 - 9y - 10xy - x^2y + 6y^2 + 2xy^2 - y^3 = 0$

The smallest possible value of $\frac{y}{x}$ is equal to $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solutions

Solution 1

Notice that $2x^3 - x^2y = x^2(2x-y)$ and $2xy^2 - y^3 = y^2(2x - y)$ . This suggests that we can factor the entire expression.

By rearranging terms, we have $\begin{align*} 18x - 9y + 4x^2 - 10xy + 6y^2 + 2x^3 - x^2 y + 2xy^2 - y^3 &= 0 \\ 9(2x-y) + (2x-y)(-2x+6y) + x^2(2x-y) + y^2(2x-y) &= 0 \\ (2x-y)(x^2 - 2x + y^2 - 6y + 9) &= 0 \end{align*}$ By the Zero Product Property, either $2x-y = 0$ or $x^2 - 2x + y^2 - 6y + 9 = 0$ . In the first case, $\frac{y}{x} = 2$ . In the second case, the equation can be rearranged into $(x-1)^2 + (y-3)^2 = 1$ , a circle with center $(1,3)$ and radius $1$ .

The values of $x$ and $y$ must be on the circle for the second case. In addition, the line connecting $(x,y)$ and the origin has a slope of $\frac{y}{k}$ . Therefore, we need to find the smallest slope of a line connecting a point on the circle to the origin. That line must be tangent to the circle.

$[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.2,xmax=3.2,ymin=-1.2,ymax=6.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); draw(circle((1,3),1)); draw((-0.75,-1)--(2.25,3),Arrows); draw((0,3)--(1,3)--(9/5,12/5),dotted); draw((1,3)--(0,0),dotted); [/asy]$ By drawing tangent lines and letting $\theta$ be half the smaller angle formed between the two tangent lines (as seen in diagram), we find that $\tan \theta = \frac13$ . Using the Double Angle Identity yields $\tan 2\theta = \frac34$ , so $\tan (90 - 2\theta) = \frac43$ .

Since the slope of the line tangent to the circle is $\frac43$ , the smallest value of $k$ is $\frac43$ , so $m+n = \boxed{7}$ .

Solution 2 (credit to NikoIsLife)

Let $y = xk$ , and we want $k$ to be minimized. Substituting for $y$ results in $\begin{align*} 18x - 4x^2 + 2x^3 - 9xk - 10x^2k - x^3k + 6x^2k^2 + 2x^3k^3 - x^3k^3 &= 0 \\ (2-k+2k^2-k^3)x^3+(-4-10k+6k^2)x^2+(18-9k)x &= 0 \\ (2-k)x((k^2 + 1)x^2 + (-6k-2)x + 9) &= 0. \end{align*}$ Since $x \ne 0$ , we must have $k = 2$ or $(k^2 + 1)x^2 + (-6k-2)x + 9 = 0$ . In the latter case, $\begin{align*} 36k^2 + 24k + 4 - 36(k^2 + 1) &\ge 0 \\ 24k - 32 &\ge 0 \\ k &\ge \frac{4}{3} \end{align*}$ The minimum value of $k$ is $\frac{4}{3}$ , so $m+n = \boxed{7}$ .

2006 iTest (Problems, Answer Key)
Preceded by: Problem U5	Followed by: Problem U7
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2006 iTest Problems/Problem U6

Contents

Problem

Solutions

Solution 1

Solution 2 (credit to NikoIsLife)

See Also