An algebra problem by Julian Yu

Algebra Level 5

Let a a , b b , and c c be real numbers such that a + b + c = 2 a+b+c=2 and a 2 + b 2 + c 2 = 12 {a}^{2}+{b}^{2}+{c}^{2}=12 .

If the difference between the maximum and minimum values of c c can be expressed as p q \dfrac{p}{q} , where p p and q q are coprime positive integers , find p + q p+q .


The answer is 19.

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2 solutions

Julian Yu
Jun 17, 2016

From the given, we have:

a + b = 2 c a+b=2-c

a 2 + b 2 = 12 c 2 {a}^{2}+{b}^{2}=12-{c}^{2}

By the Cauchy-Scwarz inequality, we have 2 ( a 2 + b 2 ) ( a + b ) 2 2({ a }^{ 2 }+{ b }^{ 2 })\ge { { (a+b) } }^{ 2 } .

Substitution of the above results gives us 3 c 2 4 c 20 = 0 { 3c }^{ 2 }-4c-20=0 , and we see that equality holds for c = 2 c=-2 and c = 10 3 c=\frac{10}{3} .

The difference between these two values is 16 3 \frac { 16 }{ 3 } so the answer is 19 \boxed { 19 } .

First solver XD

Manuel Kahayon - 4 years, 12 months ago

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Second solver!

Grant Bulaong - 4 years, 12 months ago

Here comes the third.... I thought my method would be unique and I thought of posting it first but found the same method here XD

Rishabh Jain - 4 years, 12 months ago

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I also apparently posted a very similar problem ages ago, too. What a coincidence!

Manuel Kahayon - 4 years, 12 months ago

Same solution

Manuel Kahayon - 4 years, 12 months ago

Same way... I seem to have written a solution to a similar problem before :-)

Rishabh Jain - 4 years, 12 months ago
Patrick Chatain
Jun 18, 2016

A little longer method but here it goes,

( a + b + c ) 2 = a 2 + b 2 + c 2 + 2 ( a b + b c + a c ) 4 = 12 + 2 ( a b + b c + a c ) a b + b c + a c = 4 (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac) \implies 4=12+2(ab+bc+ac) \implies ab+bc+ac=-4

Now a + b + c = 2 a + b = 2 c a+b+c=2 \implies a+b=2-c

And a b + b c + a c = 4 a b = 4 c ( a + b ) = 4 c ( 2 c ) = c 2 2 c 4 ab+bc+ac=-4 \implies ab=-4-c(a+b)=-4-c(2-c)=c^2-2c-4

So a a and b b are the real roots of the equation y = x 2 + ( c 2 ) x + ( c 2 2 c 4 ) y=x^2+(c-2)x+(c^2-2c-4)

So the discriminant of the last equation is greater than or equal to zero so ( c 2 ) 2 4 ( c 2 2 c 4 ) 0 (c-2)^2-4(c^2-2c-4)\geq0

\implies 3 c 2 + 4 c + 20 0 -3c^2+4c+20\geq0 \implies 2 c 10 3 -2\leq c\leq\frac{10}{3}

So p q = 10 3 ( 2 ) = 16 3 \frac{p}{q}=\frac{10}{3}-(-2)=\frac{16}{3} and p + q = 16 + 3 = 19 p+q=16+3=19

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