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

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m (Undo a last-resort solution, as that's a bit misleading and vague ...)
 
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== Problem ==
 
== Problem ==
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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 ==
+
== Solution 1 (Algebra: Generalized) ==
−
Simplify by repeated application of the [[difference of squares]].  
+
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>
  
−
:<math>\left((\sqrt{6} + \sqrt{7})^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - (\sqrt{6} - \sqrt{7})^2\right)</math>
+
We rewrite the original expression in terms of <math>x,y,</math> and <math>z,</math> then apply the difference of squares repeatedly:
−
:<math>= (13 + 2\sqrt{42} - 5)(5 - (13 - 2\sqrt{42}))</math>
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<cmath>\begin{align*}
−
:<math>= (2\sqrt{42} + 8)(2\sqrt{42} - 8)</math>
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(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] \\
−
:<math>= (2\sqrt{42})^2 - 8^2 = 104</math>
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&= \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>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1986|num-b=1|num-a=3}}
 
{{AIME box|year=1986|num-b=1|num-a=3}}
 +
* [[AIME Problems and Solutions]]
 +
* [[American Invitational Mathematics Examination]]
 +
* [[Mathematics competition resources]]
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

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

See also

1986 AIME (Problems • Answer Key • Resources)
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Problem 3
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