Difference between revisions of "2020 AIME II Problems/Problem 11"
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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>. | 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>. | ||
− | + | ||
− | \text{Again, by vietas, }pq=-\frac52 | + | <math>\text{Again, by vietas, }pq=-\frac52\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2}</math> |
− | \text{Multiplying everything together a} | + | <math>\text{Multiplying everything together a}\text{nd then taking the sqrt of both sides,}</math> |
− | (pqr)^2 | + | <cmath>(pqr)^2=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)</cmath> |
− | pqr | + | <cmath>pqr=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)} </cmath> |
− | \text{Now, we divide this } | + | <math>\text{Now, we divide this }\text{equation by }qr=\frac{c+2}{2} </math> |
− | \frac{pqr}{qr} | + | <cmath>\frac{pqr}{qr}=\frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}}{\frac{c+2}{2}} </cmath> |
− | p | + | <cmath>p = \frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}}{\sqrt{\frac{c+2}{2}}} </cmath> |
− | \text{Recall th}\text{at }p=\frac32 | + | <math>\text{Recall th}\text{at }p=\frac32 \text{ and square both sides}</math> |
− | \frac94 | + | <cmath>\frac94=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}} </cmath> |
− | \text{Solving gives } c=\frac{52}{19} | + | <math>\text{Solving gives } c=\frac{52}{19}, \text{ so our answer is }\boxed{071}</math> |
\end{align*} | \end{align*} | ||
Revision as of 11:29, 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 .
\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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.