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Difference between revisions of "1957 AHSME Problems/Problem 34"

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<math>\fbox{\textbf{(C)} a straight line segment not including the end-points}</math>.
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The line <math>x+y=1</math> is in [[standard form]]. Thus, it has a slope of <math>-\tfrac{1}{1}=-1</math> and a <math>y</math>-intercept at <math>(0,1)</math>. It is graphed in blue in the diagram. All the points that satisfy this system must lie on this line.
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The inequality <math>x^2+y^2<25</math> is true for all points [[strict inequality|strictly]] less than [[distance formula|distance]] <math>\sqrt{25}=5</math> from the origin. This area is a [[disk]] with radius <math>5</math>, whose open boundary is represented in red in the diagram.
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Because the <math>y</math>-intercept of the line, <math>(0,1)</math>, is inside the disk, it cannot be tangent to the disk. Thus, there must be infiniely many points along the line which are within the disk and thereby satisfy the system. However, because the inequality is strict, the segment of solutions does not include the endpoints, so our answer is <math>\fbox{\textbf{(C)} a straight line segment not including the end-points}</math>.
  
 
== See Also ==
 
== See Also ==
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{{MAA Notice}}
 
{{MAA Notice}}
 
[[Category:AHSME]][[Category:AHSME Problems]]
 
[[Category:AHSME]][[Category:AHSME Problems]]
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[[Category:Introductory Algebra Problems]]

Latest revision as of 07:25, 26 July 2024

Problem

The points that satisfy the system $x + y = 1,\, x^2 + y^2 < 25$, constitute the following set:

$\textbf{(A)}\ \text{only two points} \qquad \\ \textbf{(B)}\ \text{an arc of a circle}\qquad \\ \textbf{(C)}\ \text{a straight line segment not including the end-points}\qquad\\ \textbf{(D)}\ \text{a straight line segment including the end-points}\qquad\\ \textbf{(E)}\ \text{a single point}$

Solution

[asy] import geometry; // Axes draw((0,-8)--(0,8),arrow=Arrows); label("$y$",(0,9)); draw((-8,0)--(8,0),arrow=Arrows); label("$x$",(9,0)); // Circle and Line draw(circle((0,0),5),red+dashed); draw(line((0,1),(1,0)),blue); [/asy]

The line $x+y=1$ is in standard form. Thus, it has a slope of $-\tfrac{1}{1}=-1$ and a $y$-intercept at $(0,1)$. It is graphed in blue in the diagram. All the points that satisfy this system must lie on this line.

The inequality $x^2+y^2<25$ is true for all points strictly less than distance $\sqrt{25}=5$ from the origin. This area is a disk with radius $5$, whose open boundary is represented in red in the diagram.

Because the $y$-intercept of the line, $(0,1)$, is inside the disk, it cannot be tangent to the disk. Thus, there must be infiniely many points along the line which are within the disk and thereby satisfy the system. However, because the inequality is strict, the segment of solutions does not include the endpoints, so our answer is $\fbox{\textbf{(C)} a straight line segment not including the end-points}$.

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 33 		Followed by
Problem 35
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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