Difference between revisions of "1960 AHSME Problems/Problem 13"
Rockmanex3 (talk | contribs) (Created page with "==Problem== The polygon(s) formed by <math>y=3x+2, y=-3x+2</math>, and <math>y=-2</math>, is (are): <math>\textbf{(A) }\text{An equilateral triangle}\qquad\textbf{(B) }\text...") |
Rockmanex3 (talk | contribs) m (→See Also) |
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==Solution== | ==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 <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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Two of the side lengths are <math>\sqrt{(\frac{4}{3})^2+4^2} = \frac{4 \sqrt{10}}{3}</math> while one of the side lengths is <math>4</math>. That makes the triangle isosceles, so the answer is <math>\boxed{\textbf{(B)}}</math>. | Two of the side lengths are <math>\sqrt{(\frac{4}{3})^2+4^2} = \frac{4 \sqrt{10}}{3}</math> while one of the side lengths is <math>4</math>. That makes the triangle isosceles, so the answer is <math>\boxed{\textbf{(B)}}</math>. | ||
+ | ==See Also== | ||
+ | {{AHSME 40p box|year=1960|num-b=12|num-a=14}} | ||
− | + | [[Category:Introductory Algebra Problems]] | |
− | + | [[Category:Introductory Geometry Problems]] |
Latest revision as of 19:00, 17 May 2018
Problem
The polygon(s) formed by , and
, is (are):
Solution
The points of intersection of two of the lines are and
, so use the Distance Formula to find the sidelengths.
Two of the side lengths are while one of the side lengths is
. That makes the triangle isosceles, so the answer is
.
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |