7

In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is
$(A)$equilateral
$(B)$isosceles
$(C)$right angled
$(D)$none of these

The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$.

Using sine rule,

$a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get

$\sin^2A+\sin^2B+\sin^2C=\sin A\sin C+\sin A\sin B\sqrt3$

I am stuck here.

Vinod Kumar Punia
  • 5,648
  • 2
  • 41
  • 96
  • 1
    I'd go the other way around for eliminating possibilities:, assume the triangle is e.g. equilateral. Then $a=b=c$. Therefore $a^2+b^2+c^2 = 3a^2$ but $ac+ab\sqrt{3} = a^2(1+\sqrt{3}) \neq 3a^2$, therefore $\Delta ABC$ cannot be equilateral, e.t.c. – flawr Apr 17 '16 at 14:40
  • The answer given in my book is $ABC$ is right angled triangle,right angled at $A$ – Vinod Kumar Punia Apr 17 '16 at 14:53
  • These choices tell us at least "ABC is not right-angled isosceles triangle". – Takahiro Waki Jul 29 '18 at 20:51

4 Answers4

7

Since $0=a^2+b^2+c^2-ac-ab\sqrt{3}=\left(b-\frac{\sqrt{3}}2a\right)^2+\left(c-\frac12a\right)^2\,,$ we have $b=\frac{\sqrt{3}}2a$ and $c=\frac{1}{2}a$. Thus, $\angle A=\frac{\pi}{2}$, $\angle B=\frac{\pi}{3}$, and $\angle C=\frac{\pi}{6}$.

It can be shown that, if $\alpha,\beta,\gamma\in(0,\pi)$ and $\alpha+\beta+\gamma=\pi$, then there is a unique triangle $ABC$, up to scaling, with $BC=a$, $CA=b$, and $AB=c$ such that $$a^2+b^2+c^2=2bc\cos(\alpha)+2ca\cos(\beta)+2ab\cos(\gamma)\,.$$ To show this, one observes that the matrix $$\textbf{X}:=\begin{bmatrix}1&-\cos(\gamma)&-\cos(\beta)\\-\cos(\gamma)&1&-\cos(\alpha)\\-\cos(\beta)&-\cos(\alpha)&1\end{bmatrix}$$ is positive-semidefinite and the eigenspace associated with the eigenvalue $0$ of $\textbf{X}$ is spanned by $\begin{bmatrix}\sin(\alpha)\\\sin(\beta)\\\sin(\gamma)\end{bmatrix}$.

Batominovski
  • 49,629
6

Considering cosine law: \begin{align*} \sum_{abc} (b^2+c^2-a^2) &= \sum_{abc} 2bc\cos A \\ a^2+b^2+c^2&= 2bc\cos A+2ca\cos B+2ab\cos C \end{align*}

One possibility is $\cos A=0$, $2\cos B=1$ and $2\cos C=\sqrt{3}$.

Hence $A=90^{\circ}$, $B=60^{\circ}$ and $C=30^{\circ}$.

Proof of Uniqueness of the above solution:

Rearrange the equality as a quadratic in $b$:

$$b^2-(a\sqrt{3})b+(a^2-ac+c^2) =0 $$

The discriminant: \begin{align*} \Delta &= (-a\sqrt{3})^2-4(a^2-ac+c^2) \\ &=-a^2+4ac-4c^2 \\ &=-(a-2c)^{2} \\ &\leq 0 \end{align*}

To admit a real solution for $b$, $\Delta$ has to be zero. That is

$$a=2c$$

Now, $$3c^2-2bc\sqrt{3}+b^2=0 \implies b=c\sqrt{3}$$

Therefore, $$\fbox{$a^2=b^2+c^2$}$$

Varun Kumar
  • 327
Ng Chung Tak
  • 18,990
  • @zyx, Agree, we need to check the uniqueness also. I've just found a useful web and checking for it. http://www.artofproblemsolving.com/community/c1090h1009430 – Ng Chung Tak Apr 17 '16 at 15:16
  • @zyx, The proof is updated in my answer. – Ng Chung Tak Apr 17 '16 at 16:07
  • Your proof says that the equation is L^2 + M^2=0 where L and M are linear functions of $a,b,c$, proportional to $(a-2c)$ and $(b - c \sqrt{3})$. The equation therefore characterizes a specific shape of right triangle but not right angles at A in general (which would a^2 = b^2 + c^2). – zyx Apr 17 '16 at 16:41
  • @zyx Applying Converse of Pythagoras' Theorem or Cosine Law, we can conclude that $A$ is a right angle. – Ng Chung Tak Apr 17 '16 at 16:45
  • We already know A is 90 degrees because that is the case for the unique shape of triangle implied by the equation. However, the equation on a,b,c does not imply a^2 = b^2 + c^2 algebraically (meaning without using inequalities, which is the same as allowing complex values of a,b,c). It also does not have as its solution set all triangles with a right angle at A, which is what it would mean to "characterize right angles at A". – zyx Apr 17 '16 at 16:52
  • @zyx, Back to the question, it's talking about a triangle. Unless $a,b,c\in \mathbb{C}$ are the position of $A,B,C$ in complex plane, that was not specified in the question. Real measurements are understood. At least my answer agrees with Vinod Kumar Punia (the asker) and Batominovski (another answerer). – Ng Chung Tak Apr 17 '16 at 16:58
  • Certainly your answer proves that (C) is the correct response from the listed options. However, the problem is written in a way that suggests the equation is a characterization of right triangles, which it is not. What we learn from the solution is how the problem was composed: start from L and M and find a linear combination of L^2 and M^2 that includes $(a^2+b^2+c^2)$. The combination exists and is unique and with positive coefficients if we start from a known shape of triangle. – zyx Apr 17 '16 at 17:01
  • Of course, the coefficients of $ab,bc,ca$ can't be a crazy combination; otherwise, my attempt didn't work. Question setting usually is a kind of specialization. – Ng Chung Tak Apr 17 '16 at 17:11
5

$$a^2+b^2+c^2=ac+ab\sqrt3$$ The above equation can be re-written as $$\frac{a^2}{4}-ac+c^2+\frac{3a^2}{4}-ab\sqrt3+b^2=0$$ which is $$(\frac{a}{2}-c)^2+(\frac{\sqrt3 a}{2}-b)^2 = 0$$ which implies that $\frac{a}{2} = c$ and $\frac{\sqrt3 a}{2}= b$ .Based on this it can be concluded that the ratio of sides is 1:$\sqrt{3}$:2, which is a right angled-triangle.I hope this was helpful.

Varun Kumar
  • 327
-4

If $\triangle ABC$ is equilateral, then $3x^2=x^2(\sqrt{3}+1)$, absurd.

Now, note that the condition is symmetric respect to $b$ and $c$, ie, if $(a,b,c)$ satisfies, $(a,c,b)$ too, and then $\triangle ABC$ is isosceles with $b=c$.

Martín Vacas Vignolo
  • 4,710
  • 2
    In general, $ac \ne ab\sqrt 3$. So the condition is not symmetric w.r.t. $b$ and $c$. – TonyK Apr 17 '16 at 14:52