Difference between revisions of "1986 AIME Problems/Problem 2"
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MRENTHUSIASM (talk | contribs) (Reformatted the page and added in a generalized solution.) |
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== Problem == | == Problem == | ||
| − | Evaluate the product < | + | Evaluate the product <cmath>\left(\sqrt 5+\sqrt6+\sqrt7\right)\left(-\sqrt 5+\sqrt6+\sqrt7\right)\left(\sqrt 5-\sqrt6+\sqrt7\right)\left(\sqrt 5+\sqrt6-\sqrt7\right).</cmath> |
| − | == Solution == | + | == Solution 1 (Algebra: Specific) == |
| − | + | We repeatedly apply the difference of squares: | |
| + | <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 - (13 - 2\sqrt{42})\right) \\ | ||
| + | &= (2\sqrt{42} + 8)(2\sqrt{42} - 8) \\ | ||
| + | &= (2\sqrt{42})^2 - 8^2 \\ | ||
| + | &= \boxed{104}. | ||
| + | \end{align*}</cmath> | ||
| − | :<math>\left | + | == Solution 2 (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> The original expression becomes <cmath>(x+y+z)(x+y-z)(x-y+z)(-x+y+z),</cmath> which has cyclic symmetry with respect to <math>x,y,</math> and <math>z.</math> | |
| − | |||
| − | |||
| − | == Solution | + | We repeatedly apply the difference of squares: |
| + | <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] \\ | ||
| + | &= (2xy)^2 - (x^2+y^2-z^2)^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*}</cmath> | ||
| + | ~MRENTHUSIASM | ||
| + | |||
| + | == 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 what we are looking for. | 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 what we are looking for. | ||
Let angle <math>\theta</math> be opposite the <math>2\sqrt7</math> side. By the Law of Cosines, | Let angle <math>\theta</math> be opposite the <math>2\sqrt7</math> side. By the Law of Cosines, | ||
| − | <cmath>\cos\theta=\frac{(2\sqrt5)^2+(2\sqrt{6})^2-(2\sqrt7)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}}</cmath> | + | <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 | 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> | <cmath>\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104}</cmath> | ||
Revision as of 01:59, 7 June 2021
Contents
Problem
Evaluate the product ![\[\left(\sqrt 5+\sqrt6+\sqrt7\right)\left(-\sqrt 5+\sqrt6+\sqrt7\right)\left(\sqrt 5-\sqrt6+\sqrt7\right)\left(\sqrt 5+\sqrt6-\sqrt7\right).\]](http://latex.artofproblemsolving.com/7/a/2/7a200e0ca6ca8ce0f831a6145de8f7285aa81e49.png)
Solution 1 (Algebra: Specific)
We repeatedly apply the difference of squares:
Solution 2 (Algebra: Generalized)
More generally, let 
so that 
The original expression becomes ![\[(x+y+z)(x+y-z)(x-y+z)(-x+y+z),\]](http://latex.artofproblemsolving.com/f/1/f/f1feb14b032fb7692002454f6e08558fc3865777.png)
which has cyclic symmetry with respect to 
and 
We repeatedly apply the difference of squares:
![\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] \\ &= (2xy)^2 - (x^2+y^2-z^2)^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*}](http://latex.artofproblemsolving.com/a/4/7/a47168458d972ddfa7b3db04dd6a325d0e23f996.png)
~MRENTHUSIASM
Solution 3 (Geometry)
Notice that in a triangle with side lengths 
and 
, by Heron's formula, the area is the square root of what we are looking for.
Let angle 
be opposite the side. By the Law of Cosines,
![\[\cos\theta=\frac{\left(2\sqrt5\right)^2+\left(2\sqrt{6}\right)^2-\left(2\sqrt7\right)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}}\]](http://latex.artofproblemsolving.com/c/d/8/cd85dc76c158206235e2c14b0c63142e0e331233.png)
So 
. The area of the triangle is then
![\[\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104}\]](http://latex.artofproblemsolving.com/5/d/e/5de800c19b9b207726d6ab333c2f5f0bd61738cb.png)
So our answer is 
See also
| 1986 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
