Difference between revisions of "1960 AHSME Problems/Problem 13"

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m (See Also)
 
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==Solution==
 
==Solution==
  
[[Image:B9CCCDD9-3865-4F51-8308-230C4C1796F4.jpeg|350px]]
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<asy>import graph; size(10.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.2,xmax=4.2,ymin=-4.2,ymax=4.2;
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pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0);
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/*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1;
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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);
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Label laxis; laxis.p=fontsize(10);
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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);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
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dot((0,2),ds);
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dot((1.333,-2),ds);
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dot((-1.333,-2),ds);
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draw((0,2)--(1.333,-2)--(-1.333,-2)--(0,2));
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</asy>
  
 
The points of intersection of two of the lines are <math>(0,2)</math> and <math>(\pm \frac{4}{3} , -2)</math>, so use the Distance Formula to find the sidelengths.
 
The points of intersection of two of the lines are <math>(0,2)</math> and <math>(\pm \frac{4}{3} , -2)</math>, so use the Distance Formula to find the sidelengths.
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==See Also==
 
==See Also==
{{AHSME box|year=1960|num-b=12|num-a=14}}
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{{AHSME 40p box|year=1960|num-b=12|num-a=14}}
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[[Category:Introductory Algebra Problems]]
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[[Category:Introductory Geometry Problems]]

Latest revision as of 19:00, 17 May 2018

Problem

The polygon(s) formed by $y=3x+2, y=-3x+2$, and $y=-2$, is (are):

$\textbf{(A) }\text{An equilateral triangle}\qquad\textbf{(B) }\text{an isosceles triangle} \qquad\textbf{(C) }\text{a right triangle} \qquad \\ \textbf{(D) }\text{a triangle and a trapezoid}\qquad\textbf{(E) }\text{a quadrilateral}$

Solution

[asy]import graph; size(10.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.2,xmax=4.2,ymin=-4.2,ymax=4.2;  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);  dot((0,2),ds); dot((1.333,-2),ds); dot((-1.333,-2),ds); draw((0,2)--(1.333,-2)--(-1.333,-2)--(0,2));  [/asy]

The points of intersection of two of the lines are $(0,2)$ and $(\pm \frac{4}{3} , -2)$, so use the Distance Formula to find the sidelengths.

Two of the side lengths are $\sqrt{(\frac{4}{3})^2+4^2} = \frac{4 \sqrt{10}}{3}$ while one of the side lengths is $4$. That makes the triangle isosceles, so the answer is $\boxed{\textbf{(B)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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