Difference between revisions of "2020 AIME II Problems/Problem 11"
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==Solution 3== | ==Solution 3== | ||
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+ | We know that <math>P(x)=x^2-3x-7</math>. | ||
+ | |||
+ | Since <math>Q(0)=2</math>, the constant term in <math>Q(x)</math> is <math>2</math>. Let <math>Q(x)=x^2+ax+2</math>. | ||
+ | |||
+ | Finally, let <math>R(x)=x^2+bx+c</math>. | ||
+ | |||
+ | <math>P(x)+Q(x)=2x^2+(a-3)x-5</math>. Let its roots be <math>p</math> and <math>q</math>. | ||
+ | |||
+ | <math>P(x)+R(x)=2x^2+(b-3)x+(c-7)</math> Let its roots be <math>p</math> and <math>r</math>. | ||
+ | |||
+ | <math>Q(x)+R(x)=2x^2+(a+b)x+(c+2)</math>. Let its roots be <math>q</math> and <math>r</math>. | ||
+ | |||
+ | By vietas, <math>p+q=\frac{3-a}{2}, p+r=\frac{3-b}{2}, q+r=\frac{-(a+b)}{2}</math> | ||
+ | |||
+ | We could work out the system of equations, but it's pretty easy to see that <math>p=\frac32, q=-\frac{a}{2}, r=-\frac{b}{2}</math>. | ||
+ | \begin{align*} | ||
+ | \text{Again, by vietas, }pq=-\frac52&\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2} \\ | ||
+ | \text{Multiplying everything together a}&\text{nd then taking the sqrt of both sides,} \\ | ||
+ | (pqr)^2&=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)\\ | ||
+ | pqr&=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)} \\ | ||
+ | \text{Now, we divide this }&\text{equation by }qr=\frac{c+2}{2} \\ | ||
+ | \frac{pqr}{qr}&=\frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}}{\frac{c+2}{2}} \\ | ||
+ | p &= \frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}}{\sqrt{\frac{c+2}{2}}} \\ | ||
+ | \text{Recall th}\text{at }p=\frac32 &\text{ and square both sides}\\ | ||
+ | \frac94&=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}} \\ | ||
+ | \text{Solving gives } c=\frac{52}{19}&, \text{ so our answer is }\boxed{071} | ||
+ | \end{align*} | ||
+ | |||
+ | ~quacker88 | ||
==Video Solution== | ==Video Solution== |
Revision as of 11:20, 9 June 2020
Problem
Let , and let
and
be two quadratic polynomials also with the coefficient of
equal to
. David computes each of the three sums
,
, and
and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If
, then
, where
and
are relatively prime positive integers. Find
.
Solution 1
Let and
. We can write the following:
Let the common root of
be
;
be
; and
be
. We then have that the roots of
are
, the roots of
are
, and the roots of
are
.
By Vieta's, we have:
Subtracting from
, we get
. Adding this to
, we get
. This gives us that
from
. Substituting these values into
and
, we get
and
. Equating these values, we get
. Thus, our answer is
. ~ TopNotchMath
Solution 2
Let have shared root
,
have shared root
, and the last pair having shared root
. We will now set
, and
. We wish to find
, and now we compute
.
From here, we equate coefficients. This means
. Now,
. Finally, we know that
Solution 3
We know that .
Since , the constant term in
is
. Let
.
Finally, let .
. Let its roots be
and
.
Let its roots be
and
.
. Let its roots be
and
.
By vietas,
We could work out the system of equations, but it's pretty easy to see that .
\begin{align*}
\text{Again, by vietas, }pq=-\frac52&\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2} \\
\text{Multiplying everything together a}&\text{nd then taking the sqrt of both sides,} \\
(pqr)^2&=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)\\
pqr&=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)} \\
\text{Now, we divide this }&\text{equation by }qr=\frac{c+2}{2} \\
\frac{pqr}{qr}&=\frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}}{\frac{c+2}{2}} \\
p &= \frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}}{\sqrt{\frac{c+2}{2}}} \\
\text{Recall th}\text{at }p=\frac32 &\text{ and square both sides}\\
\frac94&=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}} \\
\text{Solving gives } c=\frac{52}{19}&, \text{ so our answer is }\boxed{071}
\end{align*}
~quacker88
Video Solution
https://youtu.be/BQlab3vjjxw ~ CNCM
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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