Difference between revisions of "1986 AIME Problems/Problem 2"
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== Problem == | == Problem == | ||
− | Evaluate the product < | + | 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) == |
− | + | 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) | + | 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> | ||
== See also == | == See also == |
Latest revision as of 00:44, 5 February 2022
Contents
Problem
Evaluate the product
Solution 1 (Algebra: Generalized)
More generally, let so that
We rewrite the original expression in terms of and
then apply the difference of squares repeatedly:
Remark
From this solution, note that the original expression has cyclic symmetry with respect to and
~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:
~Azjps (Solution)
~MRENTHUSIASM (Revision)
Solution 3 (Geometry)
Notice that in a triangle with side-lengths and
by Heron's Formula, the area is the square root of the original expression.
Let be the measure of the angle opposite the
side. By the Law of Cosines,
so
The area of the triangle is then
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.