This question comes from 1988 Irish Mathematical Olympiad, for all those interested.

A mathematical moron is given the values $b,c,\alpha$ for a triangle $ABC$ and is required to find $a$. He does this by using the cosine rule $a^2 = b^2 +c^2 −2bc\cos \alpha$ and misapplying the low of the logarithm to this to get $\log(a^2) = \log(b^2) +\log(c^2) −\log(2bc\cos \alpha)$. He proceeds to evaluate the right-hand side correctly, takes the anti-logarithms and gets the correct answer. What can be said about the triangle ABC?

Using logarithm properties, I was able to simplify the equation to $a^2=\dfrac{bc}{2\cos \alpha}$, but I can't find any significant results.

All I could say was: $a$ is not the hypotenuse of a right-angled triangle (since $\cos \alpha$ must not equal zero). Also, $\alpha$ must be less than $90$ degrees, and $a^2<b^2+c^2$.

But all that feels rather unimpressive, and I feel like I'm missing something here. I've tried replacing $a^2$ in the cosine rule, but that doesn't seem to be useful, especially since the cosine rule doesn't tell you any more about the triangle, since it is always true. Does anyone have any ideas?

(I'm not on a PC, so I can't use TeX, but if you can't understand anything above I can try clarifying it for you).

  • 1
    $\begingroup$ You can use $\text{TeX}$ anywhere you want $\endgroup$ – qwr Jul 21 '15 at 0:23
  • $\begingroup$ I tried, but it didn't come out as TeX, just the code itself. $\endgroup$ – Cataline Jul 21 '15 at 0:24
  • $\begingroup$ use dollar signs in between the text and backslash before functions $\endgroup$ – qwr Jul 21 '15 at 0:25
  • $\begingroup$ I know how to use it, but it wouldn't work. But it has been edited now using TeX, so there shouldn't be any trouble understanding the symbols. $\endgroup$ – Cataline Jul 21 '15 at 0:27
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    $\begingroup$ $$\cos \alpha = \frac{bc}{2a^2} = \frac{b^2+c^2 - a^2}{2bc} \implies a^4 - a^2(b^2+c^2)+b^2c^2 = 0 \implies (a^2-b^2)(a^2-c^2) = 0$$ $\endgroup$ – r9m Jul 21 '15 at 0:36

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