Given triangle $ABC$ with perimeter $P=25$. If $O$ is the center of inscribed circle and $CO=4$, prove that angle $C=60^0$.
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1What have you tried? – Stephen Donovan Apr 16 '21 at 12:13
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Cosine law for $c$ and Pithagorean theorem for $r$, $(p-c)$ and 4, but in this way I got 3 equations with 4 unknown values. – kmitov Apr 16 '21 at 12:16
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But you are not new to the site and you know you should add all the details of your attempt in the body of the question to avoid downvotes and votes to close. – Math Lover Apr 16 '21 at 12:19
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Yes, I know, but I did not want to give a possible wrong direction to anybody. – kmitov Apr 16 '21 at 12:27
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@kmitov pythagorean theorem for $r,\frac{p}{2}-c,4 ?$ – IITM Apr 16 '21 at 13:18
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I use $p$ to denote $p=P/2$. – kmitov Apr 16 '21 at 13:44
1 Answers
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A counterexample.
Let $c=9$, $a=8+\tfrac47\sqrt{21}$, $b=8-\tfrac47\sqrt{21}$.
Then \begin{align} \rho&=\tfrac12(a+b+c)=\frac{25}2 ,\\ S&=\tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2} =\frac{25}4\sqrt{15} ,\\ r&=\frac S\rho =\tfrac12\sqrt{15} ,\\ |IC|&=\sqrt{r^2+(\rho-c)^2} =4 . \end{align}
So, the conditions $P=25$ and $|IC|=4$ hold, but
\begin{align} \cos C&= \frac{a^2+b^2-c^2}{2ab} =\frac{17}{32} \ne\frac12 , \end{align}
hence, the angle $C\ne60^\circ$.
g.kov
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@kmitov: It's the same well-known Heron’s formula, just one of its variations, expressed in terms of $a^2,b^2,c^2$, which is sometimes more convenient than classic $\sqrt{\rho(\rho-a)(\rho-b)(\rho-c)}$. – g.kov Apr 17 '21 at 05:53
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