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I have to prove that for a triangle, $$a\cos A+b\cos B+c\cos C=\frac{8\Delta^2}{abc}$$ where $a,b,c$ are the lengths of the sides opposite to the angles A,B,C respectively. I followed the following procedure for the LHS: $$ \begin {align} a\cos A&+b\cos B+c\cos C\\ &= a\left[2\cos^2\left(\frac A2\right)-1\right]+b\left[2\cos^2\left(\frac B2\right)-1\right]+c\left[2\cos^2\left(\frac C2\right)-1\right]\\ &=a\left[2 \frac{s(s-a)}{bc} -1\right]+b\left[2 \frac{s(s-b)}{ac} -1\right]+c\left[2 \frac{s(s-c)}{ab} -1\right]\\ &=2a\left (\frac{s(s-a)}{bc}\right)+2b\left (\frac{s(s-b)}{ac}\right)+2c\left( \frac{s(s-c)}{ab}\right)-2s\\ &=\frac{2s}{abc}[a^2(s-a)+b^2(s-b)+c^2(s-c)-abc] \end{align}$$ where $$s=\frac{a+b+c}{2}$$

I want to convert that last equation into Heron's formula: $$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$

But I'm stuck there. Any help please?

Joffan
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Tejas
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4 Answers4

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From the cosine rule:

$$\cos(A)=\frac{b^2+c^2-a^2}{2bc}$$

So

$$\begin{align}a\cos(A)+b\cos(B)+c\cos(C)&=a\frac{b^2+c^2-a^2}{2bc}+b\frac{a^2+c^2-b^2}{2ac}+c\frac{a^2+b^2-c^2}{2ab}\\ &=a^2\frac{b^2+c^2-a^2}{2abc}+b^2\frac{a^2+c^2-b^2}{2abc}+c^2\frac{a^2+b^2-c^2}{2abc}\\ &=\frac{a^2b^2+a^2c^2-a^4+a^2b^2+b^2c^2-b^4+a^2c^2+b^2c^2-c^4}{2abc}\\ &=\frac{-(a-b-c)(a+b-c)(a+c-b)(a+b+c)}{2abc}\\ &=\frac{(b+c-a)(a+b-c)(a+c-b)(a+b+c)}{2abc}\\ &=\frac{(2s-2a)(2s-2c)(2s-2b)(2s)}{2abc}\\ &=\frac{8(s-a)(s-c)(s-b)(s)}{abc}\\ &=\frac{8\Delta^2}{abc}\\ \end{align}$$

Dale M
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  • That's great! :) I tried this method too. But I wasn't able to factorise the terms in your step 3. Can you tell me how you found the four factors? – Tejas Feb 05 '15 at 03:29
  • @Tejas If you want to get somewhere you can always cheat/work backwards. – Pp.. Feb 05 '15 at 03:37
  • @Pp.. Right! However, it would really help me knowing the method, in case I didn't know what lies at the other side of equality. – Tejas Feb 05 '15 at 03:43
  • @Tejas The method was: working backwards. What you were doing was more close to deducing the factorization because you took out the factor $s$ instead of multiplying it. When you do that you get a cubic polynomial (say cubic in $c$) and you could solve for $c$ and find the root $a+b$. This tells you that there is a factor of the form $c-(a+b)$. Being cyclic it tells you there are also factors $a-(b+c)$ and $b-(a+c)$. – Pp.. Feb 05 '15 at 03:55
  • To be perfectly honest - I used Wolfram Alpha to do the factorisation - in general, factorisation requires a stoke of genius or bloody hard work; I don't get that many strokes of genius and I don't like to work that hard. – Dale M Feb 05 '15 at 05:16
2

HINT:

Use Law of sines $a=2R\sin A$ etc.

Now Double-Angle Formula $2\sin A\cos A=\sin2A$

Then Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle

Finally use $\triangle=\dfrac{abc}{4R}$ (Proof)

Community
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lab bhattacharjee
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The polynomial $$\frac{a^2(s-a)+b^2(s-b)+c^2(s-c)-abc}{4}$$ vanishes for $c:=a+b$ and because it is cyclic it vanishes also at $a:=b+c$ and at $b:=a+c$. Notice that the same is true for $$(s-a)(s-b)(s-c).$$

In fact, for the second polynomial it is clear and for the first putting $c:=a+b$ in the first polynomial we get

$$\frac{2ba^2+2ab^2-2ab(a+b)}{8}\equiv 0.$$

Finally the two polynomials have the same leading coefficient $\frac{1}{8}$.

Therefore $$\frac{a^2(s-a)+b^2(s-b)+c^2(s-c)-abc}{4}=(s-a)(s-b)(s-c).$$

Pp..
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  • What is the meaning of ':='? Is it the same as '='? – Tejas Feb 05 '15 at 03:29
  • @Tejas Yes. I just put the colon to emphasize we are going to assign $a+b$ to where $c$ is in the expression. – Pp.. Feb 05 '15 at 03:36
  • But we can't have $c=a+b$ in a triangle, isn't it? – Tejas Feb 05 '15 at 03:40
  • @Tejas It doesn't matter. We are proving that two polynomials are identically equal (as opposed to equal only when the input are sides of a triangle). The idea is that you check that the two polynomials have the same roots and have the same leading coefficient. This is enough to say they are equal. It is a useful technique to remember for many proofs like this. You don't have to get dirty expanding everything. – Pp.. Feb 05 '15 at 03:43
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$t=a^2(s-a)+b^2(s-b)+c^2(s-c)-abc=a^2(s-a)+b^2(s-a)+c^2(s-a)+a(b^2+c^2-bc)-(b^3+c^3)\\b^3+c^3=(b+c)(b^2+c^2-bc)\\t=(s-a)(a^2+b^2+c^2)-(b^2+c^2-bc)(b+c-a),b+c-a=2(s-a)\\t=(s-a)(a^2+b^2+c^2-2(b^2+c^2-bc))=(s-a)(a^2-(b^2+c^2-2bc))=(s-a)(a^2-(b-c)^2)=(s-a)(a+c-b)(a+b-c)=4(s-a)(s-b)(s-c)$

chenbai
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