Maths Olympiad Prep

Library / /15 of 30

Algebra Difficulty 5.7 AIME, harder Prove it Ireland

Suppose abc0abc \neq 0. Express in terms of aa, bb, and cc, the solutions xx, yy, zz, uu, vv, ww of the equations
x+y=a,z+u=b,v+w=c,ay=bz,bu=cv,cw=ax. x + y = a, \quad z + u = b, \quad v + w = c, \quad ay = bz, \quad bu = cv, \quad cw = ax.

Solution

Label the equations 1, 2, 3, 4, 5, 6 in the order of their appearance. Using equations 3, 6 eliminate ww from the system, thereby adding to equations 1, 2, 4, 5 the equation 7: ax+cv=c2ax + cv = c^2. Next, use this and 5 to eliminate vv to produce equation 8: ax+bu=c2ax + bu = c^2. Eliminate zz from 2, 4 to produce equation 9: ay+bu=b2ay + bu = b^2. Now eliminate uu from equations 8, 9 giving axay=c2b2ax - ay = c^2 - b^2. Finally, using this in conjunction with 1, we see that
2ax=c2+a2b2,2ay=a2+b2c2. 2ax = c^2 + a^2 - b^2, \quad 2ay = a^2 + b^2 - c^2.

And so, using 6 and 4, too,
2cw=c2+a2b2,2bz=a2+b2c2. 2cw = c^2 + a^2 - b^2, \quad 2bz = a^2 + b^2 - c^2.
Hence, using equations 2, 5, we get that
2cv=2bu=2b22bz=b2+c2a2. 2cv = 2bu = 2b^2 - 2bz = b^2 + c^2 - a^2.
Thus, letting
α=b2+c2a22bc,β=c2+a2b22ca,γ=a2+b2c22ab, \alpha = \frac{b^2 + c^2 - a^2}{2bc}, \quad \beta = \frac{c^2 + a^2 - b^2}{2ca}, \quad \gamma = \frac{a^2 + b^2 - c^2}{2ab},
we see that
x=cβ,y=bγ,z=aγ,u=cα,v=bα,w=aβ. x = c\beta, \quad y = b\gamma, \quad z = a\gamma, \quad u = c\alpha, \quad v = b\alpha, \quad w = a\beta.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.