Understanding the Triangle Parameters

The problem involves a triangle ABC with sides of lengths \(a, b, c\), where \(a\) is the length of side BC, \(b\) is the length of side CA, and \(c\) is the length of side AB. We are given the perimeter \(p\) and the area \(q\) of this triangle.

• Perimeter, \(p = a + b + c\)
• Area, \(q\)

We need to evaluate the expression \(p(p-2a) \tan \left(\frac{A}{2}\right)\), where \(A\) is the angle opposite side \(a\).

Key Formulas for Triangle Calculations

To solve this problem, we will use the concept of the semi-perimeter and some standard formulas related to triangle area and angles.

• Semi-perimeter, \(s = \frac{p}{2} = \frac{a+b+c}{2}\). From this, we have \(p = 2s\).
• Heron's formula for the area of a triangle: \(q = \sqrt{s(s-a)(s-b)(s-c)}\).
• The tangent of half an angle in a triangle: \(\tan \left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\). This formula relates the half-angle to the semi-perimeter and side lengths.

Step-by-Step Evaluation of the Expression

Let's evaluate the given expression \(p(p-2a) \tan \left(\frac{A}{2}\right)\) using the formulas above.

Substitute \(p = 2s\) into the expression:

\(p(p-2a) \tan \left(\frac{A}{2}\right) = (2s)(2s-2a) \tan \left(\frac{A}{2}\right)\)

Factor out 2 from the second term:

\(= (2s) \cdot 2(s-a) \tan \left(\frac{A}{2}\right)\)

Combine the constants:

\(= 4s(s-a) \tan \left(\frac{A}{2}\right)\)

Now, substitute the formula for \(\tan \left(\frac{A}{2}\right)\):

\(= 4s(s-a) \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\)

To simplify, we can bring the terms outside the square root inside. Since \(s \gt 0\) and \(s-a \gt 0\) for a valid triangle, \(4s(s-a) = \sqrt{(4s(s-a))^2} = \sqrt{16s^2(s-a)^2}\).

So the expression becomes:

\(= \sqrt{16s^2(s-a)^2 \cdot \frac{(s-b)(s-c)}{s(s-a)}}\)

Cancel one \(s\) term and one \((s-a)\) term from the numerator and denominator inside the square root:

\(= \sqrt{16s(s-a)(s-b)(s-c)}\)

We can rewrite this as:

\(= \sqrt{16 \cdot s(s-a)(s-b)(s-c)}\)

Connecting to the Area of the Triangle

Recall Heron's formula for the area \(q\):

\(q = \sqrt{s(s-a)(s-b)(s-c)}\)

So, the expression inside the square root in our calculation is related to \(q^2\):

\(s(s-a)(s-b)(s-c) = q^2\)

Substitute this back into our simplified expression:

\(= \sqrt{16 \cdot q^2}\)

Since \(q\) represents area, \(q\) is non-negative. Therefore, \(\sqrt{q^2} = q\).

\(= 4\sqrt{q^2} = 4q\)

Thus, the expression \(p(p-2a) \tan \left(\frac{A}{2}\right)\) is equal to \(4q\).

Revision Table: Key Concepts Revisited

Additional Information on Triangle Formulas

Triangles have many interesting relationships between their sides, angles, perimeter, and area. Beyond the formulas used here, other useful formulas include:

• Area using Sine: \(q = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B = \frac{1}{2}ab \sin C\). This formula is useful when two sides and the included angle are known.
• Law of Sines: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\), where \(R\) is the circumradius of the triangle.
• Law of Cosines: \(a^2 = b^2 + c^2 - 2bc \cos A\), and similarly for \(b^2\) and \(c^2\). This is a generalization of the Pythagorean theorem and is useful for finding a side when two sides and the included angle are known, or finding an angle when all three sides are known.
• Half-Angle Cosine and Sine Formulas:
  • \(\cos \left(\frac{A}{2}\right) = \sqrt{\frac{s(s-a)}{bc}}\)
  • \(\sin \left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{bc}}\)
  These can be used to derive the tangent half-angle formula (\(\tan(A/2) = \sin(A/2) / \cos(A/2)\)).

Understanding these interconnections between different properties of a triangle is fundamental in solving geometry problems.