Latest revision as of 20:52, 19 February 2021

Let $x$ , $y$ , and $z$ be real numbers such that $x^2=3-2yz$ , $y^2=4-2xz$ , and $z^2=5-2xy$ . The value of $x^2+y^2+z^2$ can be written as $a+\frac{b}{\sqrt c}$ for positive integers $a, b, c$ , where $c$ is not divisible by the square of any prime. Find $a+b+c$ .

Solution 1

Let $n$ be our answer. Notice $(y-z)^2 = x^2 + y^2 + z^2 - (x^2 + 2yz)= n-3$ . Similarly, $(x-z)^2 = n-4$ and $(y-x)^2 = n-5$ . Now notice that since $y-z = (x-z)+(y-x)$ , we have $\sqrt{n-3}=\sqrt{n-4}+\sqrt{n-5} \implies n^2-12n+36=4n^2-36n+80$ so $3n^2-24n+44=0$ and $n=4 \pm \frac{2}{\sqrt 3}$ . The answer is $\boxed 9$ .

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 =Solution 1=
-Let <math>n</math> be our answer. Notice <math>(y-z)^2 = x^2 + y^2 + z^2 - (x2 + 2yz)= n-3</math>. Similarly, <math>(x-z)^2 = n-4</math> and <math>(y-x)^2 = n-5</math>. Now notice that since <math>y-z = (x-z)+(y-x)</math>, we have <math>\sqrt{n-3}=\sqrt{n-4}+\sqrt{n-5} \implies n^2-12n+36=4n^2-36n+80</math> so <math>3n^2-24n+44=0</math> and <math>n=4 \pm \frac{2}{\sqrt 3}</math>. The answer is <math>\boxed 9</math>.
+Let <math>n</math> be our answer. Notice <math>(y-z)^2 = x^2 + y^2 + z^2 - (x^2 + 2yz)= n-3</math>. Similarly, <math>(x-z)^2 = n-4</math> and <math>(y-x)^2 = n-5</math>. Now notice that since <math>y-z = (x-z)+(y-x)</math>, we have <math>\sqrt{n-3}=\sqrt{n-4}+\sqrt{n-5} \implies n^2-12n+36=4n^2-36n+80</math> so <math>3n^2-24n+44=0</math> and <math>n=4 \pm \frac{2}{\sqrt 3}</math>. The answer is <math>\boxed 9</math>.
 ==See also==

2019 CIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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Difference between revisions of "2019 CIME I Problems/Problem 10"

Latest revision as of 20:52, 19 February 2021

Solution 1

See also