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Difference between revisions of "1986 AIME Problems/Problem 2"

(→‎Solution 2 (Algebra: Generalized))
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&= \left[x^2+2xy+y^2-z^2\right]\left[z^2-x^2+2xy-y^2\right] \\
 
&= \left[x^2+2xy+y^2-z^2\right]\left[z^2-x^2+2xy-y^2\right] \\
 
&= \left[2xy + \left(x^2+y^2-z^2\right)\right]\left[2xy - \left(x^2+y^2-z^2\right)\right] \\
 
&= \left[2xy + \left(x^2+y^2-z^2\right)\right]\left[2xy - \left(x^2+y^2-z^2\right)\right] \\
−
&= (2xy)^2 - (x^2+y^2-z^2)^2 \\
+
&= \left(2xy\right)^2 - \left(x^2+y^2-z^2\right)^2 \\
 
&= 4x^2y^2 - x^4 - y^4 - z^4 - 2x^2y^2 + 2y^2z^2 + 2z^2x^2 \\
 
&= 4x^2y^2 - x^4 - y^4 - z^4 - 2x^2y^2 + 2y^2z^2 + 2z^2x^2 \\
 
&= 2x^2y^2 + 2y^2z^2 + 2z^2x^2 - x^4 - y^4 - z^4 \\
 
&= 2x^2y^2 + 2y^2z^2 + 2z^2x^2 - x^4 - y^4 - z^4 \\

Revision as of 02:44, 7 June 2021

Problem

Evaluate the product \[\left(\sqrt 5+\sqrt6+\sqrt7\right)\left(-\sqrt 5+\sqrt6+\sqrt7\right)\left(\sqrt 5-\sqrt6+\sqrt7\right)\left(\sqrt 5+\sqrt6-\sqrt7\right).\]

Solution 1 (Algebra: Specific)

We repeatedly apply the difference of squares: \begin{align*} \left(\left(\sqrt{6} + \sqrt{7}\right)^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - \left(\sqrt{6} - \sqrt{7}\right)^2\right) &= \left(13 + 2\sqrt{42} - 5\right)\left(5 - (13 - 2\sqrt{42})\right) \\ &= (2\sqrt{42} + 8)(2\sqrt{42} - 8) \\ &= (2\sqrt{42})^2 - 8^2 \\ &= \boxed{104}. \end{align*}

Solution 2 (Algebra: Generalized)

More generally, let $(x,y,z)=\left(\sqrt5,\sqrt6,\sqrt7\right)$ so that $\left(x^2,y^2,z^2\right)=(5,6,7).$ Note that the original expression has cyclic symmetry with respect to $x,y,$ and $z.$

We rewrite the original expression in terms of $x,y,$ and $z,$ then apply the difference of squares repeatedly: \begin{align*} (x+y+z)(x+y-z)(x-y+z)(-x+y+z) &= \left[((x+y)+z)((x+y)-z)\right]\left[((z+(x-y))(z-(x-y))\right] \\ &= \left[(x+y)^2-z^2\right]\left[z^2 - (x-y)^2\right] \\ &= \left[x^2+2xy+y^2-z^2\right]\left[z^2-x^2+2xy-y^2\right] \\ &= \left[2xy + \left(x^2+y^2-z^2\right)\right]\left[2xy - \left(x^2+y^2-z^2\right)\right] \\ &= \left(2xy\right)^2 - \left(x^2+y^2-z^2\right)^2 \\ &= 4x^2y^2 - x^4 - y^4 - z^4 - 2x^2y^2 + 2y^2z^2 + 2z^2x^2 \\ &= 2x^2y^2 + 2y^2z^2 + 2z^2x^2 - x^4 - y^4 - z^4 \\ &= 2(5)(6) + 2(6)(7) + 2(7)(5) - 5^2 - 6^2 - 7^2 \\ &= \boxed{104}. \end{align*} ~MRENTHUSIASM

Solution 3 (Geometry)

Notice that in a triangle with side lengths $2\sqrt5,2\sqrt6,$ and $2\sqrt7$, by Heron's formula, the area is the square root of what we are looking for. Let angle $\theta$ be opposite the $2\sqrt7$ side. By the Law of Cosines, \[\cos\theta=\frac{\left(2\sqrt5\right)^2+\left(2\sqrt{6}\right)^2-\left(2\sqrt7\right)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}}\] So $\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}$. The area of the triangle is then \[\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104}\]

So our answer is $\left(\sqrt{104}\right)^2=\boxed{104}$

See also

1986 AIME (Problems • Answer Key • Resources)
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Problem 3
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All AIME Problems and Solutions

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