3

In a triangle with vertices $A$, $B$, $C$, semiperimeter $s$, inradius $r$ and circumradius $R$, prove that $$\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$$ and $$\cos A+\cos B+\cos C=1+\frac rR$$

(note: we can also discover the value of $\cos A\cos B+\cos B\cos C+\cos C\cos A$ using the identity $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$)

Since the last time I've posted this question (the original thread is now deleted), I've reflected a bit on the suggestions of several users. First, I included relevant informations and defintion and second I did try to use the cosine law, but It did not give me help.

I was referred by a friend to the identities $$\begin{align} a+b+c &= 2s \tag{1} \\[4pt] ab+ac+bc &= s^2+r^2+4rR \tag{2} \\[4pt] abc &= 4Rrs \tag{3} \end{align}$$ The first and third facts are obvious, while the second I do not know for sure to be true (although it probably is) and appears to model the numerator of the first identity in $\cos A\cos B\cos C$.

Any other idea?

Quanto
  • 97,352
  • Do not delete a post, and return to post it yet again. – amWhy Feb 09 '21 at 23:48
  • Please do not delete and then repost questions. If you determine that a question needs to be improved, please use the "Edit" link under the post to make the required changes. Deleting questions and reposting them can lead to an automatic suspension of your ability to ask more questions. – Xander Henderson Feb 10 '21 at 23:21
  • @XanderHenderson I certainly understand and apologize for the reposting, however the thread, which did not have a reply, was already locked. There was no other option. –  Feb 11 '21 at 03:22
  • @na22ja The question was closed because it did not meet the standards of this site. The correct course of action, again, is to use the "Edit" link under the post to make the required changes. Once that is done, the question can be reopened. Keep in mind that deleted questions "count against you" when the automatic systems determine whether or not you are permitted to ask a question. Deleting poorly received questions is likely to lead to an automatic suspension of your ability to post new questions. – Xander Henderson Feb 11 '21 at 12:28
  • @XanderHenderson I see. Thank you for the insight. –  Feb 11 '21 at 17:13

2 Answers2

2

The area of the triangle can be expressed as $ \frac12r(a+b+c)= \frac{abc}{4R}$. Then

\begin{align} \frac rR =\frac12 \frac{\frac{abc}{R^3}}{\frac{a+b+c}R} =\frac{2\sin A \sin B \sin C}{\sin A +\sin B +\sin C}\tag1 \end{align} Note $$\sin A +\sin B +\sin C = 2\cos\frac A2\sin\frac A2 + 2\sin\frac {B+C}2\cos\frac{B-C}2\\ = 2 \cos\frac A2 (\cos\frac {B+C}2+\cos\frac{B-C}2) = 4 \cos\frac A2 \cos\frac B2 \cos\frac C2 $$

Substitute into (1) $$\frac rR = 4 \sin\frac A2 \sin\frac B2 \sin\frac C2\tag2 $$ Similarly $$\cos A +\cos B +\cos C = 1-2\sin^2\frac A2 +2\cos\frac {B+C}2\cos\frac{B-C}2\\ =1- 2 \sin\frac A2 (\cos\frac {B+C}2-\cos\frac{B-C}2)=1+ 4 \sin\frac A2 \sin\frac B2 \sin\frac C2 $$ Substitute into (2)

$$\cos A+\cos B+\cos C=1+\frac rR$$

Quanto
  • 97,352
1

Use the result $1+\frac rR=\cos A+\cos B+\cos C$ in the derivation below

\begin{align} & \frac{s^2-(2R+r)^2}{4R^2}\\ = &\frac14\left(\frac{a+b+c}{2R}\right)^2-\frac14\left(2+\frac rR\right)^2\\ =& \frac14\left[ (\sin A +\sin B +\sin C)^2 -(1+ \cos A +\cos B +\cos C)^2 \right]\\ =& \frac14\left[(\sin^2A-\cos^2 A)+(\sin^2 B-\cos^2 B) + (\sin^2 C-\cos^2 C)\right.\\ & \left.-1 -2( \cos A +\cos B +\cos C)-2(\cos (A+B)+\cos (B+C) +\cos (C+A)) \right]\\ = &\frac14( -\cos2A - \cos2B -\cos2C-1)\\ = &\frac14( -2\cos(A+B)\cos(A-B) -2\cos^2C +1 -1)\\ = &\frac12\cos C(\cos(A-B)+\cos(A+B))\\ = &\>\cos A \cos B\cos C\\ \end{align}

Quanto
  • 97,352