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If in $\triangle ABC$,$r=1$,$R=3$ and $s=5$, then find value of $a^2+b^2+c^2$

(where $r$=inradius,$R$=circumradius and $s$=semi-perimeter)

The answer given is $24$

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You know that $$r=\frac{Δ}{s}$$ Hence we get $$Δ=5$$ Also $$R=\frac{abc}{4Δ} $$ From this we get $$ abc=60$$

We also have a formula that

$$\frac{4(s-a)(s-b)(s-c)}{abc}=\frac{r}{R}$$ Hence, on substituting values we get $$(5-a)(5-b)(5-c)=5$$ Which reduces to $$125-50s+5\sum_{cyc} ab -abc = 5$$ Further on reducing we get $$\sum_{cyc} ab=38$$ Now $${(\frac{a+b+c}{2})}^2=s^2$$ $$\frac{a^2+b^2+c^2+2\sum ab}{4}=25$$ $$a^2+b^2+c^2=100-76=24$$ Hence $\sum a^2 =24$

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  • $\begingroup$ (cc'ing @Maverick) The above gives $a+b+c=10\,$, $ab+bc+ca=38\,$, $abc = 60\,$. Therefore $\,a,b,c\,$ must be the roots of the cubic $\,t^3-10t^2+38t-60=0\,$. But that cubic has one single real root, therefore no real $\,a,b,c\,$ satisfy the given conditions, and so the triangle $\,\triangle ABC\,$ doesn't in fact exist (as Michael Rozenberg's answer points out using a different argument). $\endgroup$ – dxiv Jan 2 '18 at 5:36
  • $\begingroup$ I too agree with Michael Rozenberg's answer but this question seems to formed only from the point of view of using the formulas at the correct spot to solve questions based on circles and triangles with handful of algebra. $\endgroup$ – Rohan Shinde Jan 2 '18 at 5:48
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    $\begingroup$ @dxiv Well I did not mindlessly answer this question. Before answering the question I already had read Rozenberg's answer and agreed with him. But there was no point in reinforcing this idea to the OP because Rozenberg had clearly explained why the triangle doesn't exist. So I tried the other way to answer the question which is, applying the formulas. -:) And by the way I must say, It was good sarcastic reply. -:). $\endgroup$ – Rohan Shinde Jan 2 '18 at 6:12
  • $\begingroup$ That was a link to a related "math joke", didn't mean it to come across the wrong way. FWIW one of the upvotes on this comment was mine. $\endgroup$ – dxiv Jan 2 '18 at 6:18
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$a+b+c=10$.

Let $S$ be an area of the triangle.

Thus, $$\frac{2S}{a+b+c}=1,$$ which gives $$S=5$$ or $$\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}=5$$ or $$(a+b-c)(a+c-b)(b+c-a)=40.$$ Now, by AM-GM $$40=(a+b-c)(a+c-b)(b+c-a)\leq\left(\frac{a+b-c+a+c-b+b+c-a}{3}\right)^3=\frac{1000}{27},$$ which is impossible.

Thus, this triangle does not exist.

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  • $\begingroup$ You are right. If i solve the three equations $a+b+c=10 ;ab+bc+ca=38;abc=60$ i am getting one real value and two imaginary values. $\endgroup$ – Maverick Jan 2 '18 at 6:06
  • $\begingroup$ Yes, and from here $a^2+b^2+c^2=10^2-2\cdot38=24,$ but it can not be the answer. $\endgroup$ – Michael Rozenberg Jan 2 '18 at 6:42
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I shall attempt to do a step-by-step approach. You can refer to this for one of the formulas. Hence, the area of the triangle would be $1 * 5 = 5$. Using another formula here, $$R = abc/4rs$$. Hence, $$3 = \frac{abc}{20}$$, so $abc = 60$. Since $s= 5$, perimeter $$a + b + c$$ = 10.

We have:

$$abc = 60.$$ And $$a + b + c = 10$$

$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac)$$.

Consequently, using Heron's formula, we have $$5^2= 5(5-a)(5-b)(5-c)$$, so we get $$(5-a)(5-b)(5-c) = 5$$. We have $$125 - 25a - 25b - 25c + 5ab + 5ac + 5bc -abc = 5$$. Substituting 1 and 2, we have $$125 - 250 + 5(ab + ac + bc) - 60 = 5$$. This means $$ab + ac + bc = 38$$. Substitute this into $$(a + b + c)^2= a^2 + b^2 + c^2 + 2(ab + bc + ac)$$, we get $$100 = a^2 + b^2 + c^2 + 2*38$$. Hence, $$a^2 + b^2 + c^2 = 100 - 76 = 24$$.

Note: Any helpful user please help me format my code to look like simultaneous equations. Thanks.

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  • $\begingroup$ Can you please remove the $ sign in 6 formula.And by the way your solution seems to be much similar to that of mine $\endgroup$ – Rohan Shinde Jan 2 '18 at 5:14
  • $\begingroup$ oh ok... i guess happen sometimes... but I used different formulas $\endgroup$ – QuIcKmAtHs Jan 2 '18 at 5:19
  • $\begingroup$ Yeah I noted that :-) $\endgroup$ – Rohan Shinde Jan 2 '18 at 5:21

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