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AIME 1986 · 第 2 题

AIME 1986 — Problem 2

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

Evaluate the product

(5+6+7)(5+6−7)(5−6+7)(−5+6+7).\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).
解析

Solution 1 (Algebra: Generalized)

More generally, let (x,y,z)=(5,6,7)(x,y,z)=\left(\sqrt5,\sqrt6,\sqrt7\right) so that (x2,y2,z2)=(5,6,7).\left(x^2,y^2,z^2\right)=(5,6,7).

We rewrite the original expression in terms of x,y,x,y, and z,z, then apply the difference of squares repeatedly:

(x+y+z)(x+y−z)(x−y+z)(−x+y+z)=[((x+y)+z)((x+y)−z)][((z+(x−y))(z−(x−y))]=[(x+y)2−z2][z2−(x−y)2]=[x2+2xy+y2−z2][z2−x2+2xy−y2]=[2xy+(x2+y2−z2)][2xy−(x2+y2−z2)]=(2xy)2−(x2+y2−z2)2=(2⋅5⋅6)2−(5+6−7)2=104.\begin{aligned} (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{aligned} Remark

From this solution, note that the original expression has cyclic symmetry with respect to x,y,x,y, and z:z:

(x+y+z)(x+y−z)(x−y+z)(−x+y+z)=⋯=(2xy)2−(x2+y2−z2)2=4x2y2−x4−y4−z4−2x2y2+2y2z2+2z2x2=2x2y2+2y2z2+2z2x2−x4−y4−z4.\begin{aligned} (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{aligned} ~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:

((6+7)2−52)(52−(6−7)2)=(13+242−5)(5−(13−242))=(242+8)(242−8)=(242)2−82=104.\begin{aligned} \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{aligned} ~Azjps (Solution)

~MRENTHUSIASM (Revision)

Solution 3 (Geometry)

Notice that in a triangle with side-lengths 25,26,2\sqrt5,2\sqrt6, and 27,2\sqrt7, by Heron's Formula, the area is the square root of the original expression.

Let θ\theta be the measure of the angle opposite the 272\sqrt7 side. By the Law of Cosines,

cos⁡θ=(25)2+(26)2−(27)22⋅25⋅26=16830=215,\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⁡θ=1−cos⁡2θ=1315.\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}.

The area of the triangle is then

sin⁡θ2⋅25⋅26=2630⋅230=104,\frac{\sin\theta}{2}\cdot 2\sqrt5\cdot 2\sqrt6=\sqrt{\frac{26}{30}}\cdot 2\sqrt{30}=\sqrt{104}, so our answer is (104)2=104.\left(\sqrt{104}\right)^2=\boxed{104}.