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Difference between revisions of "1969 AHSME Problems/Problem 31"

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== Solution ==
 
== Solution ==
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<math>\fbox{}</math>
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Each point on the square can be in the form <math>(0,y)</math>, <math>(1,y)</math>, <math>(x,0)</math>, and <math>(x,1)</math>, where <math>0 \le x,y \le 1</math>.  Making the appropriate substitutions results in points being <math>(-y^2,0)</math>, <math>(1-y^2,2y)</math>, <math>(x^2,0)</math>, and <math>(x^2 - 1,2x)</math> on the <math>uv</math>-plane. 
  
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== See also ==
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Notice that since <math>v \ge 0</math>, none of the points are below the u-axis, so options A,B, and E are out.  Since <math>x = \tfrac{v}{2}</math>, <math>u = (\tfrac{v}{2})^2 - 1</math>, so <math>v = 2\sqrt{u+1}</math>, where <math>-1 \le u \le 0</math>.  That means some of the lines are not straight, so the answer is <math>\boxed{\textbf{(D)}}</math>.
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== See Also ==
 
{{AHSME 35p box|year=1969|num-b=30|num-a=32}}   
 
{{AHSME 35p box|year=1969|num-b=30|num-a=32}}   
  
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[[Category: Intermediate Geometry Problems]]
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[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:24, 20 June 2018

Problem

Let $OABC$ be a unit square in the $xy$-plane with $O(0,0),A(1,0),B(1,1)$ and $C(0,1)$. Let $u=x^2-y^2$, and $v=xy$ be a transformation of the $xy$-plane into the $uv$-plane. The transform (or image) of the square is:

[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,2)--(1,0)--(0,-2)--cycle,dot); MP("(A)",(-5,2),SW); MP("O",(0,0),SW); MP("(-1,0)",(-1,0),SW); MP("(0,2)",(0,2),NE); MP("(1,0)",(1,0),SE); MP("(0,-2)",(0,-2),SE); [/asy]

[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw(arc((1.5,0),2.5,126,234),black); draw(arc((-1.5,0),2.5,54,-54),black); MP("(B)",(-5,2),SW); MP("O",(0,0),SW); MP("(-1,0)",(-1,0),SW); MP("(0,2)",(0,2),NE); MP("(1,0)",(1,0),SE); MP("(0,-2)",(0,-2),SE); [/asy]


[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,2)--(1,0),black); MP("(C)",(-5,2),SW); MP("O",(0,0),SW); MP("(-1,0)",(-1,0),SW); MP("(0,2)",(0,2),NE); MP("(1,0)",(1,0),SE); [/asy]


[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw(arc((1.5,0),2.5,126,180),black); draw(arc((-1.5,0),2.5,54,0),black); MP("(D)",(-5,2),SW); MP("O",(0,0),SW); MP("(-1,0)",(-1,0),SW); MP("(0,2)",(0,2),NE); MP("(1,0)",(1,0),SE); MP("(0,-2)",(0,-2),SE); [/asy]


[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,dot); MP("(E)",(-5,2),SW); MP("O",(.1,.1),SW); MP("(-1,0)",(-1,0),SW); MP("(0,1)",(0,1),NE); MP("(1,0)",(1,0),SE); MP("(0,-1)",(0,-1),SE); [/asy]

Solution

Each point on the square can be in the form $(0,y)$, $(1,y)$, $(x,0)$, and $(x,1)$, where $0 \le x,y \le 1$. Making the appropriate substitutions results in points being $(-y^2,0)$, $(1-y^2,2y)$, $(x^2,0)$, and $(x^2 - 1,2x)$ on the $uv$-plane.

Notice that since $v \ge 0$, none of the points are below the u-axis, so options A,B, and E are out. Since $x = \tfrac{v}{2}$, $u = (\tfrac{v}{2})^2 - 1$, so $v = 2\sqrt{u+1}$, where $-1 \le u \le 0$. That means some of the lines are not straight, so the answer is $\boxed{\textbf{(D)}}$.

See Also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 30 		Followed by
Problem 32
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
All AHSME Problems and Solutions

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