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

Latest revision as of 23:44, 4 February 2022

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 $\left(x^2,y^2,z^2\right)=(5,6,7).$

We rewrite the original expression in terms of $x,y,$ and $z,$ then apply the difference of squares repeatedly: $\begin{align*} (x+y+z)(x+y-z)(x-y+z)(-x+y+z) &= \left[((x+y)+z)((x+y)-z)\right]\left[((z+(x-y))(z-(x-y))\right] \\ &= \left[(x+y)^2-z^2\right]\left[z^2 - (x-y)^2\right] \\ &= \left[x^2+2xy+y^2-z^2\right]\left[z^2-x^2+2xy-y^2\right] \\ &= \left[2xy + \left(x^2+y^2-z^2\right)\right]\left[2xy - \left(x^2+y^2-z^2\right)\right] \\ &= \left(2xy\right)^2 - \left(x^2+y^2-z^2\right)^2 \\ &= \left(2\cdot\sqrt5\cdot\sqrt6\right)^2 - \left(5+6-7\right)^2 \\ &= \boxed{104}. \end{align*}$ Remark

From this solution, note that the original expression has cyclic symmetry with respect to $x,y,$ and $z:$ $\begin{align*} (x+y+z)(x+y-z)(x-y+z)(-x+y+z) &= \cdots \\ &= \left(2xy\right)^2 - \left(x^2+y^2-z^2\right)^2 \\ &= 4x^2y^2 - x^4 - y^4 - z^4 - 2x^2y^2 + 2y^2z^2 + 2z^2x^2 \\ &= 2x^2y^2 + 2y^2z^2 + 2z^2x^2 - x^4 - y^4 - z^4. \end{align*}$ ~MRENTHUSIASM

Solution 2 (Algebra: Specific)

We group the first and 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 the original expression.

Let $\theta$ be the measure of the angle 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 1: / Line 1: @@
 == Problem ==
-Evaluate the product <math>(\sqrt 5+\sqrt6+\sqrt7)(-\sqrt 5+\sqrt6+\sqrt7)(\sqrt 5-\sqrt6+\sqrt7)(\sqrt 5+\sqrt6-\sqrt7)</math>.
+Evaluate the product <cmath>\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).</cmath>
+== Solution 1 (Algebra: Generalized) ==
+More generally, let <math>(x,y,z)=\left(\sqrt5,\sqrt6,\sqrt7\right)</math> so that <math>\left(x^2,y^2,z^2\right)=(5,6,7).</math>
+We rewrite the original expression in terms of <math>x,y,</math> and <math>z,</math> then apply the difference of squares repeatedly:
+<cmath>\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 \\
+&= \left(2\cdot\sqrt5\cdot\sqrt6\right)^2 - \left(5+6-7\right)^2 \\
+&= \boxed{104}.
+\end{align*}</cmath>
+<u><b>Remark</b></u>
+From this solution, note that the original expression has cyclic symmetry with respect to <math>x,y,</math> and <math>z:</math>
+<cmath>\begin{align*}
+(x+y+z)(x+y-z)(x-y+z)(-x+y+z) &= \cdots \\
+&= \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.
+\end{align*}</cmath>
+~MRENTHUSIASM
+== Solution 2 (Algebra: Specific) ==
+We group the first and 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) \\
+&= \left(2\sqrt{42} + 8\right)\left(2\sqrt{42} - 8\right) \\
+&= \left(2\sqrt{42}\right)^2 - 8^2 \\
+&= \boxed{104}.
+\end{align*}</cmath>
+~Azjps (Solution)
+~MRENTHUSIASM (Revision)
+== Solution 3 (Geometry) ==
+Notice that in a triangle with side-lengths <math>2\sqrt5,2\sqrt6,</math> and <math>2\sqrt7,</math> by Heron's Formula, the area is the square root of the original expression.
+Let <math>\theta</math> be the measure of the angle opposite the <math>2\sqrt7</math> side. By the Law of Cosines,
+<cmath>\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}},</cmath>
+so <math>\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}.</math>
+The area of the triangle is then
+<cmath>\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104},</cmath>
+so our answer is <math>\left(\sqrt{104}\right)^2=\boxed{104}.</math>
-== Solution ==
-{{solution}}
 == See also ==
-* [[1986 AIME Problems]]
+{{AIME box|year=1986|num-b=1|num-a=3}}
+* [[AIME Problems and Solutions]]
+* [[American Invitational Mathematics Examination]]
+* [[Mathematics competition resources]]
-{{AIME box|year=1986|num-b=1|num-a=3}}
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

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