Difference between revisions of "1984 AHSME Problems/Problem 29"
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Therefore, we must have <math> 3-2\sqrt{2}\leq k\leq3+2\sqrt{2} </math>, and the maximum value of <math> k=\frac{y}{x} </math> is <math> {3+2\sqrt{2}}, \boxed{\text{A}} </math>. | Therefore, we must have <math> 3-2\sqrt{2}\leq k\leq3+2\sqrt{2} </math>, and the maximum value of <math> k=\frac{y}{x} </math> is <math> {3+2\sqrt{2}}, \boxed{\text{A}} </math>. | ||
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+ | ==Solution 2== | ||
+ | The equation represents a circle of radius <math>\sqrt{6}</math> centered at <math>A=(3,3)</math>. To find the maximal <math>k</math> with <math>y=kx</math> is equivalent to finding the maximum slope of a line passing through the origin <math>O</math> and intersecting the circle. The steepest such line is tangent to the circle at some point <math>X</math>. We have <math>XA=\sqrt{6}</math>, <math>OA=\sqrt{18}</math>, <math>\angle OXA = 90</math> because the line is tangent to the circle. Using the pythagorean theorem, we have <math>OX=\sqrt{12}</math>. | ||
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+ | The slope we are looking for is equivalent to <math>\tan (\theta + 45)</math> where <math>\angle AOX = \theta</math>. Using tangent addition, | ||
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+ | <cmath> \tan (\theta + 45)= \frac{\tan \theta + \tan 45}{1-\tan\theta\tan 45} = \frac{\frac{1}{\sqrt{2}}+1}{1-\frac{1}{\sqrt{2}}}=3+2\sqrt{3}</cmath> | ||
+ | |||
+ | So <math>\boxed{A}</math> is the answer | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1984|num-b=28|num-a=30}} | {{AHSME box|year=1984|num-b=28|num-a=30}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:58, 27 January 2019
Contents
Problem
Find the largest value for for pairs of real numbers which satisfy .
Solution
Let , so that . Substituting this into the given equation yields . Multiplying this out and forming it into a quadratic yields .
We want to be a real number, so we must have the discriminant . The discriminant is . Therefore, we must have , or . The roots of this quadratic, using the quadratic formula, are , so the quadratic can be factored as . We can now separate this into cases:
Case 1: Then, both terms in the factored quadratic are negative, so the inequality doesn't hold.
Case 2: Then, the first term is positive and the second is negative and the second is positive, so the inequality holds.
Case 3: Then, both terms are positive, so the inequality doesn't hold.
Also, when or , the equality holds.
Therefore, we must have , and the maximum value of is .
Solution 2
The equation represents a circle of radius centered at . To find the maximal with is equivalent to finding the maximum slope of a line passing through the origin and intersecting the circle. The steepest such line is tangent to the circle at some point . We have , , because the line is tangent to the circle. Using the pythagorean theorem, we have .
The slope we are looking for is equivalent to where . Using tangent addition,
So is the answer
See Also
1984 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.