4

If in a triangle $ABC$,$c$ is the longest side and $a^2+b^2=2Rc$,where $R$ is the circumradius of the triangle.Then prove that $ABC$ is a right triangle.

$a^2+b^2=2Rc\Rightarrow a^2+b^2=\frac{c^2}{\sin C}$
$\sin C=\frac{c^2}{a^2+b^2}$,how to proceed ahead?I am stuck.

diya
  • 3,589
  • Does the question exclude the use of web resources such as Wolfram Alpha (e.g. is it a contest question)? Are you familiar with multivariable calculus? – Marconius Aug 18 '15 at 18:26

2 Answers2

4

In a $\triangle ABC$, $c$ is the longest side and $a^2+b^2=2Rc$, where $R$ is the circumradius of the triangle. Prove that $\triangle ABC$ is a right triangle.

We have \begin{align} \sin C&=\frac{c^2}{a^2+b^2}, \\ \cos C&= \tfrac12 \frac{a^2+b^2-c^2}{ab}, \end{align} hence \begin{align} 1&=\left(\frac{c^2}{a^2+b^2}\right)^2 +\left(\tfrac12 \frac{-c^2+a^2+b^2}{ab}\right)^2 \end{align} $\Rightarrow$ \begin{align} 0&= \frac{(a^2+b^2-c^2) ((6 a^2 b^2+a^4+b^4) c^2+b^4 a^2+b^2 a^4-a^6-b^6) }{ 4 a^2 b^2 (a^2+b^2)^2 } \end{align}

So one solution for $c$ in terms of $a$ and $b$ certainly corresponds to a right triangle, $c^2=a^2+b^2$. Assuming $a>b$, the other one is

\begin{align} c^2&= \frac{ (a^2+b^2)(a^2-b^2)^2 }{ (a^2+b^2)^2+4 a^2 b^2 } \end{align} or, \begin{align} c^2&=a^2 \frac{ \left(1-\tfrac{b^2}{a^2}\right)^2 }{ 1+4\tfrac{b^2}{a^2+b^2}+\tfrac{b^2}{a^2} } <a^2, \end{align} hence for this solution $c$ is not the longest side, the answer is: it is indeed a right triangle.

Edit

Example solution with "wrong" (that is, not right at all) triangle for which $a^2+b^2=2Rc$ ($c<a$):

enter image description here

\begin{align} a&=2,\ b=1,\ c=\tfrac3{41}\sqrt{205}\approx 1.047645443 \\ \rho&=\tfrac12(a+b+c)=\tfrac32+\tfrac3{82}\sqrt{205} \\ S&=\sqrt{\rho(\rho-a)(\rho-b)(\rho-c)}=\tfrac9{41} \\ R&=\frac{abc}{4S}=\tfrac16\sqrt{205} \end{align} Check: \begin{align} a^2+b^2&=5 \\ 2Rc&=2\cdot \tfrac16\sqrt{205}\cdot \tfrac3{41}\sqrt{205}=5. \end{align}

g.kov
  • 13,581
  • +1 - That's a very "wrong" triangle indeed - in fact, to claim this satisfies the requirements would be really "obtuse". – Marconius Aug 18 '15 at 21:16
  • @Marconius: Thanks. Yes, they all are obtuse, also one type among them is an isosceles one. – g.kov Aug 19 '15 at 01:57
1

You are correct to use the sine law.

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \tag{1}$$

If you apply the sine law to sides $a,b$ the main equation becomes:

$$\begin{align} &4R^2 \sin^2A + 4R^2 \sin^2B = 4R^2 \sin C \implies \\ &\sin^2A + \sin^2B = \sin C\tag{2} \end{align}$$

For a triangle $C = 180^\circ - (A+B)$, so then

$$\sin C = \sin(A+B)$$

Substituting this into (2), we get:

$$\sin^2A + \sin^2B = \sin (A+B) \tag{3}$$

Can you show that a solution is $A+B=90^\circ$, so that $C=90^\circ$?

Marconius
  • 5,635
  • There is an obvious continuation from: $\sin^2A + \sin^2B = \sin (A+B) $? –  Aug 18 '15 at 17:33
  • 1
    It's fairly straightforward to show that it is a solution. I'm not sure how to show that it is the only solution for which $0 < A + B < 120^\circ$, other than to consider a function such as $f(x,y)=\sin^2x+\sin^2y-\sin(x+y)$ and find the extrema $\to$ this may be overkill for the problem. – Marconius Aug 18 '15 at 17:49
  • You right, only I'm beginning to doubt this problem is correct. The condition (3) looks too weak to give only one solution ... but I might be wrong. –  Aug 18 '15 at 17:55
  • WolframAlpha seems to indicate that this might be the only solution for $C>A,B$, e.g. https://www.wolframalpha.com/input/?i=plot+sin%5E2%28x%29+%2B+sin%5E2%28y%29+-+sin%28x%2By%29 . (I think the red zones are where the function is negative). Despite its apparent simplicity (and symmetry), eqn (3) has a lot of hidden venom! – Marconius Aug 18 '15 at 18:20
  • https://www.wolframalpha.com/input/?i=plot+sin%5E2%28x%29+%2B+sin%5E2%28y%29+-+sin%28x%2By%29%3D0%2Cx%3E%3D0%2Cy%3E%3D0%2Cx%2By%3C%3Dpi# – Booldy Aug 18 '15 at 18:46
  • $a^2+b^2=c^2/ \sin C$ implies ab (the sides) so AB (the angles) (re comment by Marconius) – DanielWainfleet Aug 18 '15 at 20:06
  • Surely if $c$ is the longest side, then $C$ is the largest angle. $a^2+b^2=c^2/ \sin C$ just implies that $c^2\le(a^2+b^2)$. – Marconius Aug 18 '15 at 21:11
  • @Marconius,how can we prove that $\sin^2 A+\sin^2 B=\sin(A+B)$ gives $A+B=\frac{\pi}{2}?$ – diya Aug 19 '15 at 10:31
  • $A+B=\frac{\pi}{2}$ is a solution, because then $\sin^2A+\sin^2B=\sin^2A+\sin^2(\frac{\pi}{2}-A)=\sin^2A+\cos^2A=1$ and $\sin(A+B)=\sin(\frac{\pi}{2})=1$. To prove that it is the only class of solutions for $A,B\in(0,\frac{\pi}{2})$ is much more difficult. You can't use a trig identity because there are other solutions when $A,B$ are not restricted to $(0,\frac{\pi}{2})$. Maybe QM-AM could work, but things could get really messy. – Marconius Aug 19 '15 at 16:07