Difference between revisions of "2020 AIME II Problems/Problem 11"
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<cmath>Q + R = 2x^2 + (a + b)x + (c + 2)</cmath> | <cmath>Q + R = 2x^2 + (a + b)x + (c + 2)</cmath> | ||
Let the common root of <math>P+Q,P+R</math> be <math>r</math>; <math>P+R,Q+R</math> be <math>s</math>; and <math>P+Q,Q+R</math> be <math>t</math>. We then have that the roots of <math>P+Q</math> are <math>r,t</math>, the roots of <math>P + R</math> are <math>r, s</math>, and the roots of <math>Q + R</math> are <math>s,t</math>. | Let the common root of <math>P+Q,P+R</math> be <math>r</math>; <math>P+R,Q+R</math> be <math>s</math>; and <math>P+Q,Q+R</math> be <math>t</math>. We then have that the roots of <math>P+Q</math> are <math>r,t</math>, the roots of <math>P + R</math> are <math>r, s</math>, and the roots of <math>Q + R</math> are <math>s,t</math>. | ||
+ | |||
+ | By Vieta's, we have: | ||
+ | <cmath>(1) r + t = \dfrac{3 - a}{2}</cmath> | ||
+ | <cmath>(2) r + s = \dfrac{3 - b}{2}</cmath> | ||
+ | <cmath>(3) s + t = \dfrac{-a - b}{2}</cmath> | ||
+ | <cmath>(4) rt = \dfrac{-5}{2}</cmath> | ||
+ | <cmath>(5) rs = \dfrac{c - 7}{2}</cmath> | ||
+ | <cmath>(6) st = \dfrac{c + 2}{2}</cmath> | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/BQlab3vjjxw ~ CNCM | https://youtu.be/BQlab3vjjxw ~ CNCM | ||
==See Also== | ==See Also== |
Revision as of 19:28, 7 June 2020
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
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:
Video Solution
https://youtu.be/BQlab3vjjxw ~ CNCM