1961 AHSME Problems/Problem 40

Problem

Find the minimum value of $\sqrt{x^2+y^2}$ if $5x+12y=60$ .

$\textbf{(A)}\ \frac{60}{13}\qquad \textbf{(B)}\ \frac{13}{5}\qquad \textbf{(C)}\ \frac{13}{12}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$

Solutions

Solution 1

Let $x^2 + y^2 = r^2$ , so $r = \sqrt{x^2 + y^2}$ . Thus, this problem is really finding the shortest distance from the origin to the line $5x + 12y = 60$ .

$[asy]import graph; size(8.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=13.2,ymin=-5.2,ymax=6.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); draw((0,5)--(12,0),Arrows); draw(Circle((0,0),60/13),dotted); [/asy]$

From the graph, the shortest distance from the origin to the line is the altitude to the hypotenuse of the right triangle with legs $5$ and $12$ . The hypotenuse is $13$ and the area is $30$ , so the altitude to the hypotenuse is $\frac{60}{13}$ , which is answer choice $\boxed{\textbf{(A)}}$ .

Solution 2

Solve for $y$ in the linear equation. $12y = 60 - 5x$ $y = 5 - \frac{5x}{12}$ Substitute $y$ in $\sqrt{x^2+y^2}$ . $\sqrt{x^2 + (5 - \frac{5x}{12})^2}$ $\sqrt{x^2 + 25 - \frac{25x}{6} + \frac{25x^2}{144}}$ $\sqrt{\frac{169x^2}{144} - \frac{25x}{6} + 25}$ To find the minimum, find the vertex of the quadratic. The x-value of the vertex is $\frac{25}{6} \cdot \frac{72}{169} = \frac{300}{169}$ . Thus, the minimum value is $\sqrt{\frac{169}{144} \cdot \frac{300^2}{169^2} - \frac{25}{6} \cdot \frac{300}{169} + 25}$ $\sqrt{\frac{10000}{16 \cdot 169} - \frac{1250}{169} + 25}$ $\sqrt{\frac{625}{169} - \frac{1250}{169} + \frac{4225}{169}}$ $\sqrt{\frac{3600}{169}}$ $\frac{60}{13}$ The answer is $\boxed{\textbf{(A)}}$ .

Solution 3

By Cauchy-Schwarz, $(5^2 + 12^2)(x^2 + y^2) >= (5x+12y)^2$ Therefore: $(13^2)(x^2 + y^2) >= 60^2$ $13\sqrt{x^2 + y^2} >= 60$ Therefore: $\sqrt{x^2 + y^2} >= 60/13$

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 39	Followed by Last Question
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