Determining the Nature of Triangle ABC from a Side Relation

The problem provides a relationship between the sides of a triangle ABC, given by the equation:

\(a^2 + b^2 + c^2 = ac + \sqrt{3} bc\)

We need to determine if the triangle is equilateral, isosceles, right-angled, or scalene but not right-angled based on this equation.

Let's rearrange the given equation:

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

To identify the nature of the triangle, we can try to manipulate this equation into a form that relates to known triangle properties, such as the Pythagorean theorem (\(a^2 + b^2 = c^2\)) or the cosine rule (\(c^2 = a^2 + b^2 - 2ab \cos C\)).

Let's multiply the entire equation by 2 to potentially reveal sum of squares forms:

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

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

Now, let's try to group terms to form perfect squares. We look for patterns like \((x-y)^2 = x^2 - 2xy + y^2\). The terms \(-2ac\) and \(-2\sqrt{3} bc\) are suggestive.

Consider grouping terms involving \(a\) and \(c\), and terms involving \(b\) and \(c\):

• Terms with \(a\) and \(c\): \(2a^2 - 2ac + 2c^2\). Can we form \((a-kc)^2\)? We need \(a^2 - 2akc + k^2c^2\). If \(k=1\), we get \((a-c)^2 = a^2 - 2ac + c^2\). We have \(2a^2, -2ac, 2c^2\).
• Terms with \(b\) and \(c\): \(2b^2 - 2\sqrt{3} bc + 2c^2\). Can we form \((b-mc)^2\)? We need \(b^2 - 2mbc + m^2c^2\). If \(m=\sqrt{3}\), we get \((b-\sqrt{3}c)^2 = b^2 - 2\sqrt{3} bc + 3c^2\). We have \(2b^2, -2\sqrt{3} bc, 2c^2\).

Let's try a different grouping for the sum of squares, aiming for the form \((X)^2 + (Y)^2 = 0\).

Consider the terms \(a^2 - ac\) and \(b^2 - \sqrt{3} bc\). These look like parts of squares involving \(c\).

• For \(a^2 - ac\), we could complete the square with a term involving \(c^2\): \(a^2 - ac + (\frac{c}{2})^2 = (a - \frac{c}{2})^2 = a^2 - ac + \frac{c^2}{4}\).
• For \(b^2 - \sqrt{3} bc\), we could complete the square with a term involving \(c^2\): \(b^2 - \sqrt{3} bc + (\frac{\sqrt{3}c}{2})^2 = (b - \frac{\sqrt{3}c}{2})^2 = b^2 - \sqrt{3} bc + \frac{3c^2}{4}\).

Let's see if the sum of these two squares uses up all terms in the original equation \(a^2 + b^2 + c^2 - ac - \sqrt{3} bc = 0\).

\((a - \frac{c}{2})^2 + (b - \frac{\sqrt{3}c}{2})^2 = (a^2 - ac + \frac{c^2}{4}) + (b^2 - \sqrt{3} bc + \frac{3c^2}{4})\)

\(= a^2 - ac + \frac{c^2}{4} + b^2 - \sqrt{3} bc + \frac{3c^2}{4}\)

\(= a^2 + b^2 + (\frac{c^2}{4} + \frac{3c^2}{4}) - ac - \sqrt{3} bc\)

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

This matches the left side of the given equation! So, the equation \(a^2 + b^2 + c^2 = ac + \sqrt{3} bc\) is equivalent to:

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

Since the square of a real number is always non-negative, the sum of two squares can be zero only if both squares are zero.

Thus, we must have:

1. \(a - \frac{c}{2} = 0 \implies a = \frac{c}{2}\)
2. \(b - \frac{\sqrt{3}c}{2} = 0 \implies b = \frac{\sqrt{3}c}{2}\)

We have found relationships between the sides \(a, b,\) and \(c\). Let's check if these relationships satisfy the Pythagorean theorem, which would indicate a right-angled triangle.

Substitute the expressions for \(a\) and \(b\) in terms of \(c\) into the Pythagorean equation \(a^2 + b^2 = c^2\):

\((\frac{c}{2})^2 + (\frac{\sqrt{3}c}{2})^2 = \frac{c^2}{4} + \frac{3c^2}{4} = \frac{c^2 + 3c^2}{4} = \frac{4c^2}{4} = c^2\)

So, we have \(a^2 + b^2 = c^2\). This is the Pythagorean theorem, which means the triangle is right-angled with the right angle at vertex \(C\) (opposite side \(c\)).

We can also determine the angles using the side ratios. Since \(a = \frac{1}{2}c\) and \(b = \frac{\sqrt{3}}{2}c\), we can consider a right triangle with hypotenuse \(c\). \(\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{c} = \frac{c/2}{c} = \frac{1}{2}\). Thus, angle \(A = 30^\circ\). \(\sin B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{b}{c} = \frac{\sqrt{3}c/2}{c} = \frac{\sqrt{3}}{2}\). Thus, angle \(B = 60^\circ\). The third angle \(C = 180^\circ - (A + B) = 180^\circ - (30^\circ + 60^\circ) = 180^\circ - 90^\circ = 90^\circ\).

Since one angle is \(90^\circ\), the triangle is a right-angled triangle.

Let's consider the side lengths. The sides are in the ratio \(a : b : c = \frac{1}{2}c : \frac{\sqrt{3}}{2}c : c\). Multiplying by 2, the ratio is \(1 : \sqrt{3} : 2\). Since \(1 \neq \sqrt{3} \neq 2\), the sides are all of different lengths. Therefore, the triangle is also a scalene triangle. However, the options provide classifications that might overlap. The most specific nature indicated by \(a^2+b^2=c^2\) is that it is a right-angled triangle.

Comparing with the options:

• Equilateral: All sides equal. Not true (sides are \(c/2, \sqrt{3}c/2, c\)).
• Isosceles: At least two sides equal. Not true (sides are all different lengths).
• Right angled triangle: One angle is 90 degrees. True (angle \(C = 90^\circ\)).
• Scalene but not right angled: All sides different, and no angle is 90 degrees. This triangle is scalene, but it IS right-angled, so this option is incorrect.

The most accurate classification from the given options is "Right angled triangle".

Revision Table: Triangle Nature from Side Equations

Additional Information: Classifying Triangles

Triangles can be classified based on their side lengths or their angles.

Classification by Side Lengths:

• Equilateral Triangle: All three sides are equal in length. All three angles are equal (\(60^\circ\) each).
• Isosceles Triangle: At least two sides are equal in length. The angles opposite the equal sides are also equal.
• Scalene Triangle: All three sides have different lengths. All three angles are different.

Classification by Angles:

• Acute Triangle: All three angles are acute (less than \(90^\circ\)).
• Right Triangle (or Right-angled Triangle): One angle is a right angle (\(90^\circ\)). The side opposite the right angle is called the hypotenuse, and the other two sides are called legs. The Pythagorean theorem (\(a^2 + b^2 = c^2\)) applies to right triangles.
• Obtuse Triangle: One angle is obtuse (greater than \(90^\circ\)).

A triangle can fit multiple classifications, e.g., a right isosceles triangle (angles \(45^\circ, 45^\circ, 90^\circ\)) or a scalene right triangle (angles like \(30^\circ, 60^\circ, 90^\circ\) or other combinations). The triangle in this problem is both scalene and right-angled.