Difference between revisions of "2020 AIME II Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | Let <math>P( | + | Let <math>P(x) = x^2 - 3x - 7</math>, and let <math>Q(x)</math> and <math>R(x)</math> be two quadratic polynomials also with the coefficient of <math>x^2</math> equal to <math>1</math>. David computes each of the three sums <math>P + Q</math>, <math>P + R</math>, and <math>Q + R</math> and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If <math>Q(0) = 2</math>, then <math>R(0) = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. |
− | ==Solution== | + | ==Solution 1== |
Let <math>Q(x) = x^2 + ax + 2</math> and <math>R(x) = x^2 + bx + c</math>. We can write the following: | Let <math>Q(x) = x^2 + ax + 2</math> and <math>R(x) = x^2 + bx + c</math>. We can write the following: | ||
<cmath>P + Q = 2x^2 + (a - 3)x - 5</cmath> | <cmath>P + Q = 2x^2 + (a - 3)x - 5</cmath> | ||
Line 11: | Line 11: | ||
By Vieta's, we have: | By Vieta's, we have: | ||
− | <cmath> | + | <cmath> r + t = \dfrac{3 - a}{2}\tag{1}</cmath> |
− | <cmath> | + | <cmath>r + s = \dfrac{3 - b}{2}\tag{2}</cmath> |
− | <cmath> | + | <cmath>s + t = \dfrac{-a - b}{2}\tag{3}</cmath> |
− | <cmath> | + | <cmath>rt = \dfrac{-5}{2}\tag{4}</cmath> |
− | <cmath> | + | <cmath>rs = \dfrac{c - 7}{2}\tag{5}</cmath> |
− | <cmath> | + | <cmath>st = \dfrac{c + 2}{2}\tag{6}</cmath> |
− | Subtracting <math>(3)</math> from <math>(1)</math>, we get <math>r - s = \dfrac{3 + b}{2}</math>. Adding this to <math>(2)</math>, we get <math>2r = 3 \implies r \dfrac{3}{2}</math>. This gives us that <math>t = \dfrac{-5}{3}</math> from <math>(4)</math>. Substituting these values into <math>(5)</math> and <math>(6)</math>, we get <math>s = \dfrac{c-7}{3}</math> and <math>s = \dfrac{-3c - 6}{10}</math>. Equating these values, we get <math>\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)</math>. Thus, our answer is <math>52 + 19 = \boxed{071}</math>. | + | Subtracting <math>(3)</math> from <math>(1)</math>, we get <math>r - s = \dfrac{3 + b}{2}</math>. Adding this to <math>(2)</math>, we get <math>2r = 3 \implies r = \dfrac{3}{2}</math>. This gives us that <math>t = \dfrac{-5}{3}</math> from <math>(4)</math>. Substituting these values into <math>(5)</math> and <math>(6)</math>, we get <math>s = \dfrac{c-7}{3}</math> and <math>s = \dfrac{-3c - 6}{10}</math>. Equating these values, we get <math>\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)</math>. Thus, our answer is <math>52 + 19 = \boxed{071}</math>. ~ TopNotchMath |
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | <math>\text{Again, by vietas, }pq=-\frac52\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2}\text{, } \text{multiplying everything together a}\text{nd taking the sqrt of both sides,}</math> | ||
+ | <cmath>(pqr)^2=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)</cmath> | ||
+ | <cmath>pqr=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)} </cmath> | ||
+ | <math>\text{Dividing this }\text{equation by }qr=\frac{c+2}{2} </math> | ||
+ | <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> | ||
+ | <cmath>p = \frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}}{\sqrt{\frac{c+2}{2}}} </cmath> | ||
+ | <math>\text{Recall th}\text{at }p=\frac32 \text{ and square both sides}</math> | ||
+ | <cmath>\frac94=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}} </cmath> | ||
+ | <math>\text{Solving gives } c=\frac{52}{19}, \text{ so our answer is }\boxed{071}</math> | ||
+ | |||
+ | ~quacker88 | ||
+ | |||
+ | ==Solution 3 (Official MAA)== | ||
+ | Let the common root of <math>P+Q</math> and <math>P+R</math> be <math>p</math>, the common root of <math>P+Q</math> and <math>Q+R</math> be <math>q</math>, and the common root of <math>Q+R</math> and <math>P+R</math> be <math>r</math>. Because <math>p</math> and <math>q</math> are both roots of <math>P+Q</math> and <math>P+Q</math> has leading coefficient <math>2</math>, it follows that <math>P(x) + Q(x) = 2(x-p)(x-q).</math> Similarly, <math>P(x) + R(x) = 2(x-p)(x-r)</math> and <math>Q(x) + R(x) = 2(x-q)(x-r)</math>. Adding these three equations together and dividing by <math>2</math> yields<cmath>P(x) + Q(x) + R(x) = (x-p)(x-q) + (x-p)(x-r) + (x-q)(x-r),</cmath>so | ||
+ | <cmath>P(x) = (P(x) + Q(x) + R(x)) - (Q(x) + R(x))</cmath> | ||
+ | <cmath>= (x-p)(x-q) + (x-p)(x-r) - (x-q)(x-r) </cmath> | ||
+ | <cmath>= x^2 - 2px + (pq + pr - qr).</cmath> | ||
+ | Similarly, | ||
+ | <cmath>Q(x) = x^2 - 2qx + (pq + qr - pr) \text{~ and}</cmath> | ||
+ | <cmath>R(x) = x^2 - 2rx + (pr + qr - pq).</cmath> | ||
+ | Comparing the <math>x</math> coefficients yields <math>p = \tfrac32</math>, and comparing the constant coefficients yields <math>-7 = pq + pr - qr = \tfrac32(q+r) - qr</math>. The fact that <math>Q(0) = 2</math> implies that <math>\tfrac32(q-r) + qr = 2</math>. Adding these two equations yields <math>q = -\tfrac53</math>, and so substituting back in to solve for <math>r</math> gives <math>r=-\tfrac{27}{19}</math>. Finally,<cmath>R(0) = pr + qr - pq = \left(-\frac{27}{19}\right)\left(\frac32-\frac53\right) + \frac52 = \frac{9}{38} + \frac52 = \frac{52}{19}.</cmath>The requested sum is <math>52 + 19 = 71</math>. Note that <math>Q(x) = x^2 + \frac{10}3x + 2</math> and <math>R(x) = x^2 + \frac{54}{19}x + \frac{52}{19}</math>. | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/BQlab3vjjxw ~ CNCM | https://youtu.be/BQlab3vjjxw ~ CNCM | ||
+ | |||
+ | Another one: | ||
+ | |||
+ | https://www.youtube.com/watch?v=AXN9x51KzNI | ||
+ | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=II|num-b=10|num-a=12}} | {{AIME box|year=2020|n=II|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} | ||
+ | |||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
− |
Latest revision as of 21:16, 28 December 2023
Contents
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
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 .
~quacker88
Solution 3 (Official MAA)
Let the common root of and
be
, the common root of
and
be
, and the common root of
and
be
. Because
and
are both roots of
and
has leading coefficient
, it follows that
Similarly,
and
. Adding these three equations together and dividing by
yields
so
Similarly,
Comparing the
coefficients yields
, and comparing the constant coefficients yields
. The fact that
implies that
. Adding these two equations yields
, and so substituting back in to solve for
gives
. Finally,
The requested sum is
. Note that
and
.
Video Solution
https://youtu.be/BQlab3vjjxw ~ CNCM
Another one:
https://www.youtube.com/watch?v=AXN9x51KzNI
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.