Difference between revisions of "1965 AHSME Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | <math>\fbox{E}</math> | + | |
| + | <asy> | ||
| + | |||
| + | draw((-16,0)--(16,0), arrow=Arrows); | ||
| + | label("$x$",(17.5,0)); | ||
| + | draw((0,-16)--(0,16), arrow=Arrows); | ||
| + | label("$y$", (0,17.5)); | ||
| + | |||
| + | draw(circle((0,0),8),red); | ||
| + | draw((-14,-12/5)--(14, 72/5), blue, arrow=Arrows); | ||
| + | |||
| + | </asy> | ||
| + | |||
| + | For an ordered pair <math>(x,y)</math> to satisfy the restrictions, the point <math>(x,y)</math> must lie on the graph of the line <math>5y-3x=15</math> (shown in blue in the diagram) and on the closed [[disk]] given by <math>x^2+y^2 \leq 16</math> (whose boundary is red in the diagram). Because the <math>y</math>-intercept of the line, <math>(0,3)</math>, is within the disk (and not on the boundary of the disk), the line cannot be tangent to the disk, and so the disk must contain infinitely many of the points on the line. Thus, our answer is <math>\fbox{\textbf{(E)} greater than any finite number}</math> | ||
==See Also== | ==See Also== | ||
{{AHSME 40p box|year=1965|num-b=12|num-a=14}} | {{AHSME 40p box|year=1965|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Latest revision as of 16:02, 18 July 2024
Problem
Let 
be the number of number-pairs 
which satisfy 
and 
. Then is:
Solution
![[asy] draw((-16,0)--(16,0), arrow=Arrows); label("$x$",(17.5,0)); draw((0,-16)--(0,16), arrow=Arrows); label("$y$", (0,17.5)); draw(circle((0,0),8),red); draw((-14,-12/5)--(14, 72/5), blue, arrow=Arrows); [/asy]](http://latex.artofproblemsolving.com/a/a/f/aafe973592865590f58a9b6a59ade0f5d5a00018.png)
For an ordered pair to satisfy the restrictions, the point must lie on the graph of the line 
(shown in blue in the diagram) and on the closed disk given by 
(whose boundary is red in the diagram). Because the 
-intercept of the line, 
, is within the disk (and not on the boundary of the disk), the line cannot be tangent to the disk, and so the disk must contain infinitely many of the points on the line. Thus, our answer is 
See Also
| 1965 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
