Difference between revisions of "1984 AHSME Problems/Problem 29"
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So <math>\boxed{A}</math> is the answer | So <math>\boxed{A}</math> is the answer | ||
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+ | ==Solution 3== | ||
+ | |||
+ | Following the steps of Solution 2, we get <math>OX=\sqrt{12}</math>. Let <math>X = (x, y)</math>. Based on the distance to the origin and to the centre of the circle | ||
+ | |||
+ | <cmath>x^2 + y^2 = 12</cmath> | ||
+ | |||
+ | <cmath>(x - 3)^2 + (y - 3)^2 = 6</cmath> | ||
+ | |||
+ | Expanding the latter and substituting the former into the equation gives <math>30 - 6(x + y) = 6</math>. <math>x + y = 4</math>. We can square this and remove the 1st equation to get <math>2xy = 4</math>, or <math>xy = 2</math>. | ||
+ | |||
+ | Substitute <math>y = 4 - x</math> to get | ||
+ | |||
+ | <cmath>x(4 - x) = 2</cmath> | ||
+ | |||
+ | <cmath>x^2 - 4x + 2 = 0</cmath> | ||
+ | |||
+ | <cmath>x = 2 \pm \sqrt{2}</cmath> | ||
+ | |||
+ | Since <math>x</math> and <math>y</math> are symmetric, we want <math>x</math> to be the lower number to maximize <math>k</math>, so <math>x = 2 - \sqrt{2}</math> and <math>y = 2 + \sqrt{2}</math>. | ||
+ | |||
+ | <cmath>\frac{y}{x} = \frac{2 + \sqrt{2}}{2 - \sqrt{2}}</cmath> | ||
+ | |||
+ | <cmath>\frac{y}{x} = \frac{(2 + \sqrt{2})^2}{2}</cmath> | ||
+ | |||
+ | <cmath>\frac{y}{x} = 3 + 2\sqrt{2}</cmath> | ||
==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}} |
Latest revision as of 20:09, 12 September 2022
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
Solution 3
Following the steps of Solution 2, we get . Let
. Based on the distance to the origin and to the centre of the circle
Expanding the latter and substituting the former into the equation gives .
. We can square this and remove the 1st equation to get
, or
.
Substitute to get
Since and
are symmetric, we want
to be the lower number to maximize
, so
and
.
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.