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I need help with this easy triangle problem:

We know:

  • One of the sides a = 16 cm.
  • The inradius r = 6cm.
  • And the circumradius R = 17 cm. That's all.

We must find the lengths of the other two sides.

I've got the values of the sin and cos of alpha (8/17 and 15/17, respectively) and yet I don't know how to make the system of equations.

I know this problem is far easier from all the others here (at least in comparison with what I have seen), but I am really stuck and need some assistance.

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    $\begingroup$ Do you know that $(ABC)=sr=\frac{abc}{4R}$? $\endgroup$
    – RobinSparrow
    Commented Jul 7 at 16:53
  • $\begingroup$ Yeah but I don't know what can I get from it... I got b + c = 4/51(bc) - 16. $\endgroup$
    – Georgi Angelov
    Commented Jul 7 at 16:56
  • $\begingroup$ What is $\alpha$? $\endgroup$
    – Kolakoski54
    Commented Jul 7 at 16:56
  • $\begingroup$ Let $\Delta$ be the area of the triangle and $s$ the semi-perimeter. Try using $R=\frac{abc}{4\Delta}$ and $r=\frac \Delta s$. You can use Heron's formula for relating $a,b,c$ to $\Delta$. This is indeed a painful problem (I wouldn't recommend anyone doing this). For more formulas for $R$ and $r$ check out SL Loney's Trigonometry. $\endgroup$
    – xyz1234
    Commented Jul 7 at 16:56
  • $\begingroup$ We don't know the value of alpha, just it's sin from 2R = a/sin(alpha). $\endgroup$
    – Georgi Angelov
    Commented Jul 7 at 16:57

1 Answer 1

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$$\tan (\frac x2) = \frac{\sin (x)}{1+\cos(x)} \implies \tan (\frac{\alpha}2) = \frac{8}{17+15} = \frac1{4}$$

$$\tan(\frac{\alpha}2) = \frac r{s-a} \iff s-16 = 4 \cdot 6 \iff s = 40$$

$$a+b+c = 80 \implies b+c = 64$$

$$a^2 = b^2+c^2-2bc \cos (\alpha) \iff 256 = 64^2-2bc - 2bc \frac{15}{17}$$

$$2bc(\frac{32}{17}) = 64^2-16^2 = 80 \cdot 48 =16^2 \cdot 15 \iff bc = 60 \cdot 17$$

$$z^2 - 64z + 60 \cdot 17 = 0 \implies (z-32)^2 = 32^2-60 \cdot 17 = 4$$

$$z = 32 \pm 2 \implies b,c \in \{30,34\}$$

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  • $\begingroup$ Thank you very much!!! Just to clarify, tan(x/2)=sin(x)/(1+cos(x)) and tan(α/2)=r/(s−a) are formulas right? And if so, where can I find more formulas like these? $\endgroup$
    – Georgi Angelov
    Commented Jul 7 at 17:32
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    $\begingroup$ yes those are famous identities, I'm not quite sure where there is a list of all of them, they are many $\endgroup$
    – hellofriends
    Commented Jul 7 at 18:11

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