Difference between revisions of "2010 AMC 12A Problems/Problem 21"
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Notice that squaring the first equation yields <math>p^2+q^2+r^2+2pq+2qr+2pr= 25</math>, which is similar to the second equation. | Notice that squaring the first equation yields <math>p^2+q^2+r^2+2pq+2qr+2pr= 25</math>, which is similar to the second equation. | ||
− | Subtracting this from the second equation, we get <math>2pq+2pr+2qr = 4</math>. Now that we have | + | Subtracting this from the second equation, we get <math>2pq+2pr+2qr = 4</math>. Now that we have the <math>pq+pr+qr</math> term, we can manpulate the equations to |
yield the sum of squares. <math>2(p^2+q^2+r^2+2pq+2qr+2pr)-2pq-2pr-2qr= 25*2-4</math> or <math>2p^2+2q^2+2r^2+2pq+2qr+2pr = 46</math>. We finally reach <math>(p+q)^2+(q+r)^2+(p+r)^2 = 46</math>. | yield the sum of squares. <math>2(p^2+q^2+r^2+2pq+2qr+2pr)-2pq-2pr-2qr= 25*2-4</math> or <math>2p^2+2q^2+2r^2+2pq+2qr+2pr = 46</math>. We finally reach <math>(p+q)^2+(q+r)^2+(p+r)^2 = 46</math>. | ||
Revision as of 12:26, 24 November 2018
Contents
[hide]Problem
The graph of lies above the line
except at three values of
, where the graph and the line intersect. What is the largest of these values?
Solution 1
The values in which
intersect at
are the same as the zeros of
.
Since there are zeros and the function is never negative, all
zeros must be double roots because the function's degree is
.
Suppose we let ,
, and
be the roots of this function, and let
be the cubic polynomial with roots
,
, and
.
In order to find we must first expand out the terms of
.
[Quick note: Since we don't know ,
, and
, we really don't even need the last 3 terms of the expansion.]
All that's left is to find the largest root of .
Solution 2
The values in which
intersect at
are the same as the zeros of
.
We also know that this graph has 3 places tangent to the x-axis, which means that each root has to have a multiplicity of 2.
Let the function be
.
Applying Vieta's formulas, we get or
.
Applying it again, we get, after simplification,
.
Notice that squaring the first equation yields , which is similar to the second equation.
Subtracting this from the second equation, we get . Now that we have the
term, we can manpulate the equations to
yield the sum of squares.
or
. We finally reach
.
Since the answer choices are integers, we can guess and check squares to get in some order. We can check that this works by adding then and seeing
. We just need to take the lowest value in the set, square root it, and subtract the resulting value from 5 to get
.
Note: One could also multiply by 2 and subtract from
to obtain
The ordered triple {16,4,1} sums to 21, and the answer choices are all positive integers, therefore the answer is 4.
Alternative method:
After reaching and
, we can algebraically derive
.
Applying Vieta's formulas on the term yields
.
Notice that , so
Subtracting this from yields
, so
, which means that
,
, and
are the roots of the cubic
, and it is not hard to find that these roots are
,
, and
. The largest of these values is
.
See also
2010 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.