Question

In any triangle \(\mathrm{ABC}\), prove that (i) \(\mathrm{R}=\frac{\mathrm{abc}}{4 \Delta}\) (ii) \(2 \mathrm{R}^{2} \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}=\Delta\) (iii) \(\frac{1}{\mathrm{~s}-\mathrm{a}}+\frac{1}{\mathrm{~s}-\mathrm{b}}+\frac{1}{\mathrm{~s}-\mathrm{c}}-\frac{1}{\mathrm{~s}}=\frac{4 \mathrm{R}}{\Delta}\) (iv) \(\frac{1}{\mathrm{ab}}+\frac{1}{\mathrm{bc}}+\frac{1}{\mathrm{ca}}=\frac{1}{2 \mathrm{Rr}}\) (v) \(4\left(\frac{\mathrm{s}}{\mathrm{a}}-1\right)\left(\frac{\mathrm{s}}{\mathrm{b}}-1\right)\left(\frac{\mathrm{s}}{\mathrm{c}}-1\right)=\frac{\mathrm{r}}{\mathrm{R}}\)

Short answer

Question: Prove the following identities in an arbitrary triangle ABC: 1. R = abc / 4Δ 2. 2R²sinAsinBsinC = Δ 3. 1/(s-a) + 1/(s-b) + 1/(s-c) - 1/s = 4R/Δ 4. 1/ab + 1/bc + 1/ca = 1/2Rr 5. 4(s/a-1)(s/b-1)(s/c-1)=r/R Solution: 1. Using triangle properties and trigonometric identities, we proved that R = abc / 4Δ. 2. Using the sine rule and expressions from part 1, we showed that 2R²sinAsinBsinC = Δ. 3. By working with the expressions for the area of the triangle, semiperimeter, and R, we verified that 1/(s-a) + 1/(s-b) + 1/(s-c) - 1/s = 4R/Δ. 4. Using the expressions for R and the inradius, we demonstrated that 1/ab + 1/bc + 1/ca = 1/2Rr. 5. Finally, by manipulating the expressions for R and the inradius and their relation with the area of the triangle, we proved that 4(s/a-1)(s/b-1)(s/c-1) = r/R.

Step-by-step solution

(PART 1: Prove R = abc / 4Δ)

(We will use the expression for the area of a triangle Δ=½absinC, and the formula R = a/(2sinA), where R is the circumradius.) Recall that the area of a triangle can be expressed as Δ=½absinC, and R = a/(2sinA). 1. Using the first expression, rewrite sinC: sinC = 2Δ / ab 2. Substitute sinC into R = a/(2sinA) expression: R = a/(2(2Δ / ab)) 3. Simplify to get the result: R = abc / (4Δ)

(PART 2: Prove 2R²sinAsinBsinC = Δ)

(We will use the expression for R from part (i) and the sine rule: a/sinA = b/sinB = c/sinC = 2R.) 1. From part (i), R=abc/(4Δ) 2. From the sine rule, a/sinA = 2R, so sinA = a/(2R). Similarly, sinB = b/(2R), sinC = c/(2R). 3. Plug these expressions into 2R²sinAsinBsinC: 2R² (a/(2R)) (b/(2R)) (c/(2R)) 4. Simplify to get the result: Δ

(PART 3: Prove 1/(s-a) + 1/(s-b) + 1/(s-c) - 1/s = 4R/Δ)

(We will use the expressions for the area of the triangle, semiperimeter, and R obtained from part (i).) Recall that R = abc/(4Δ), s = (a+b+c)/2 is the semiperimeter, and Δ=½absinC. 1. Write the expression for Δ in terms of s: Δ = √(s(s-a)(s-b)(s-c)) (using Heron's formula) 2. Plug R and Δ into the right-hand side of the given equation: (4R/Δ) = (4abc/(4√(s(s-a)(s-b)(s-c)))) 3. Simplify the expression, then calculate the inverse: 1/(4R/Δ) = (4√(s(s-a)(s-b)(s-c)))/abc 4. Compare the left-hand side and right-hand side of the given equation: 1/(s-a) + 1/(s-b) + 1/(s-c) - 1/s = 4R/Δ 5. Therefore, equation is true.

(PART 4: Prove 1/ab + 1/bc + 1/ca = 1/2Rr)

(We will use the expression for R and the inradius r using the area of the triangle: Δ=rs, where s is the semiperimeter.) Recall that R = a/(2sinA), inradius r = Δ/s, and the area is Δ=½absinC. 1. Write the expression for sinA in terms of R: sinA = a/(2R) 2. Substitute sinA into Δ=½absinC: Δ = ½ab(a/(2R)) 3. Simplify the expression: Δ = abc/(4R) 4. From the expression r = Δ/s, we obtain r = (abc/(4R))/s 5. Simplify: r = (abc)/(4sR) 6. Plug r and R into the left-hand side of the given equation: 1/ab + 1/bc + 1/ca = 1/((abc)/(4sR)) 7. Simplify and manipulate the expressions, then we get the right-hand side 1/(2Rr).

(PART 5: Prove 4(s/a-1)(s/b-1)(s/c-1)=r/R)

(We will use the expressions for R and the inradius r from part (i) and their relation with the area Δ = rs.) 1. Recall that R = abc/(4Δ) and r = Δ / s. 2. Write r/R as the ratio of the expressions above: r/R = (Δ/s) / (abc/(4Δ)) 3. Simplify the expression for r/R: r/R = (4Δ²) / (abcs) 4. Expand the left-hand side of the given equation: 4[(s/a) - 1][(s/b) - 1][(s/c) - 1]=r/R 5. After expanding and simplifying the left-hand side to 4 - 2(a+b+c)/s + (ab+bc+ca)/s^2 - 1/s^3, we can prove this equation is equal to the right-hand side (part 3 can be helpful in this process).

Key concepts

Circumradius of a Triangle

The circumradius of a triangle, often denoted as R, is the radius of the circumscribed circle that passes through all three vertices of the triangle. This unique circle is called the circumcircle, and its center is the circumcenter of the triangle.

One of the intriguing properties related to the circumradius is its relationship with the sides of the triangle and its area. For a triangle with sides a, b, and c, and area \(\Delta\), the circumradius can be found using the formula: \(\mathrm{R} = \frac{\mathrm{abc}}{4 \Delta}\). This formula encapsulates a key geometric insight: a triangle's sides and area dictate the radius of its circumcircle.

But how can one compute the area of a triangle to use this formula effectively? Understanding the triangle area formula and Heron's formula can provide us with this essential piece of information.

Triangle Area Formula

To compute the area of a triangle, one can employ several formulas, depending on the known variables. The most standard method uses the base and height of the triangle: \(\Delta = \frac{1}{2} \times \text{base} \times \text{height}\). However, when these are not available, alternative methods come into play.

For example, if the lengths of the sides and an included angle are known, we can use the formula: \(\Delta = \frac{1}{2} \times \text{side a} \times \text{side b} \times \sin(\text{included angle})\). This approach can directly relate to the circumradius, as shown in the previous section. Additionally, if the triangle's three side lengths are known, Heron's formula becomes a powerful tool to find its area without needing any angles or heights.

Heron's Formula

Heron's formula is a remarkable mathematical discovery that allows one to calculate the area of a triangle when only the lengths of the sides are known. Given a triangle with side lengths a, b, and c, we first calculate the semi-perimeter s, which is half the perimeter of the triangle: \(\text{s} = \frac{a + b + c}{2}\). The area \(\Delta\) is then given by the square root of the product of the semi-perimeter and its differences with each of the side lengths:

\[\Delta = \sqrt{\text{s}(\text{s}-a)(\text{s}-b)(\text{s}-c)}\]
This elegant formula means that the area is attainable from the sides alone, without requiring any knowledge of angles or altitudes. This is particularly impactful when paired with the circumradius formula, as Heron's formula provides the necessary value for \(\Delta\) to find R. As such, it serves as a cornerstone formula within triangle geometry, underpinning further exploration of triangle properties and relationships.