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

Revision as of 23:38, 9 June 2021

Problem

Evaluate the product $\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).$

Solution 1 (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 2 (Algebra: Specific)

We group the first and the last factors as well as the two middle factors, then apply the difference of squares repeatedly: $\begin{align*} \left(\left(\sqrt{6} + \sqrt{7}\right)^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - \left(\sqrt{6} - \sqrt{7}\right)^2\right) &= \left(13 + 2\sqrt{42} - 5\right)\left(5 - \left(13 - 2\sqrt{42}\right)\right) \\ &= \left(2\sqrt{42} + 8\right)\left(2\sqrt{42} - 8\right) \\ &= \left(2\sqrt{42}\right)^2 - 8^2 \\ &= \boxed{104}. \end{align*}$ ~Azjps (Solution)

~MRENTHUSIASM (Revision)

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 $\left(\sqrt{104}\right)^2=\boxed{104}$

@@ Line 20: / Line 20: @@
 == Solution 2 (Algebra: Specific) ==
-We apply the difference of squares repeatedly:
+We group the first and the last factors as well as the two middle factors, then apply the difference of squares repeatedly:
 <cmath>\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 - \left(13 - 2\sqrt{42}\right)\right) \\

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