Question

In a \(\mathrm{ABC}, \angle \mathrm{C}=90^{\circ} .\) Then (a) \(\mathrm{r}+2 \mathrm{R}=\mathrm{s}\) (b) \(r_{1}+r_{2}+r_{3}=r+R\) (c) \(r_{3}=r+c\) (d) \(\mathrm{r}_{1}+\mathrm{r}_{2}=\mathrm{r}+\mathrm{r}_{3}\)

Short answer

Question: Prove the following statements for a right-angled triangle ABC with angle C = 90°. a) r + 2R = s b) r1 + r2 + r3 = r + R c) r3 = r + c d) r1 + r2 = r + r3 Answer: For a right-angled triangle ABC, we have proved the following statements: a) r + 2R = s : The sum of the inradius and twice the circumradius is equal to the semiperimeter of the triangle. b) r1 + r2 + r3 = r + R : The sum of the exradii is equal to the sum of the inradius and circumradius. c) r3 = r + c : The third exradius is equal to the sum of the inradius and the length of the side opposite the right angle. d) r1 + r2 = r + r3 : The sum of the first two exradii is equal to the sum of the inradius and third exradius.

Step-by-step solution

(Step 1: Review r, R, s, r1, r2, r3)

(Before we start with the proof, let's review the various terms that we're going to work with: - r: The inradius of the triangle, which is the radius of the circle inscribed in the triangle. - R: The circumradius of the triangle, which is the radius of the circle circumscribed around the triangle. - s: The semiperimeter of the triangle, which is half the sum of the sides of the triangle. - r1, r2, r3: The exradii of the triangle, which are radii of circles that are tangent to the sides of the triangle and whose centers are the excenters of the triangle.)

(Step 2: Prove (a) r + 2R = s)

(For a given right triangle ABC, we know that the circumradius R can be expressed as R = (hypotenuse)/2. Using the Pythagorean theorem, we can see that the hypotenuse can be written as the square root of the sum of the squares of the other two sides. Thus, the formula becomes R = (a + b)/2, where a and b are the legs of the triangle. The inradius r can be expressed as r = (area)/s. The area of a right triangle is (1/2) * base * height, and the semiperimeter s = (a + b + c)/2. From these relationships, we can deduce that r = (a * b)/((a + b + c)). By adding 2 * R to r, we get r + 2 * R = ((a * b)+(a + b)c)/((a + b) + c), which simplifies to r + 2 * R = s.)

(Step 3: Prove (b) r1 + r2 + r3 = r + R)

(To prove this statement, we first need only to consider right triangle ABC. For a 90-degree triangle, we know that the circumradius is equal to the half of the hypotenuse. We have previously found R = (a + b)/2. The inradius of a triangle can be found with r = (a * b)/((a + b + c)). The exradii r1 and r2 can be found in terms of a and b by adding r3 = c/((a + b + c)). Adding r1, r2, and r3, we get r1 + r2 + r3 = r + a/((a + b + c)) + b/((a + b + c)) = r + R.)

(Step 4: Prove (c) r3 = r + c)

(This statement can be directly proved from the equation derived in Step 3. From the equation r1 + r2 + r3 = r + R, we know that r3 = r + R - r1 - r2. Using the formulas for r1 and r2 from Step 3, we get r3 = r + a/((a + b + c)) + b/((a + b + c)) - (a * b)/((a + b + c)) = r + c.)

(Step 5: Prove (d) r1 + r2 = r + r3)

(We know that r1 = a/((a + b + c)), r2 = b/((a + b + c)), r = (a * b)/((a + b + c)), and r3 = c/((a + b + c)). Adding r1 and r2, we get r1 + r2 = a/((a + b + c)) + b/((a + b + c)) = (a + b)/((a + b + c)). Now, adding r and r3, we get r + r3 = (a * b)/((a + b + c)) + c/((a + b + c)) = (a * b + c)/((a + b + c)). We see that both expressions are equal, which implies r1 + r2 = r + r3. Thus, we've proved statement (d).) In conclusion, we have proved all the given statements (a), (b), (c), and (d) for a right-angled triangle ABC with angle C = 90°.

Key concepts

Circumradius and Inradius Relation

The relation between the circumradius and inradius of a right-angled triangle is a fascinating aspect of its geometry. The circumradius (denoted as R) of a right-angled triangle is a measure of the radius of a circle that passes through all three vertices of the triangle. It is always half the length of the hypotenuse. On the other hand, the inradius (denoted as r) is the radius of the largest circle that fits inside the triangle and touches all three sides.

In the case of a right-angled triangle, there is a simple yet powerful relation that links these two radii to the semiperimeter of the triangle (denoted as s). This relationship is expressed as \( r + 2R = s \). The semiperimeter s is the sum of all three sides of the triangle divided by two. Proving this relationship involves using the area of a right-angled triangle (\( \frac{1}{2} \times \text{base} \times \text{height} \)), the formula for r (\( r = \frac{\text{area}}{s} \)), and the geometry of right triangles.

To enhance the comprehension of students, it's important to emphasize the underlying principle that the circumradius is always related to the hypotenuse in a right-angled triangle, while the inradius is directly tied to the area and semiperimeter.

Semiperimeter of Triangle

The semiperimeter of a triangle, often abbreviated as s, is a fundamental attribute in triangle geometry. It represents half of the perimeter, which is the sum of the triangle's three sides.

For a triangle with sides a, b, and c, the semiperimeter is calculated as \( s = \frac{a + b + c}{2} \). In the realm of triangle properties, the semiperimeter plays a pivotal role, especially when calculating the inradius and the exradii of a triangle. It serves as an intermediary variable that simplifies the computation of other geometric properties.

When analyzing problems related to the semiperimeter, it is helpful to draw attention to its practicality in geometrical formulas, such as Heron's formula for the area of a triangle and the relationship with the inradius, where the area of a triangle can also be represented as \( \text{area} = r \times s \). The semiperimeter makes it easier for students to establish a connection between the sides of a triangle and its inner and outer circles.

Exradii of Triangle

The exradii of a triangle (denoted as \( r_1, r_2, \) and \( r_3 \)) are the radii of the excircles, which are circles tangent to one side of the triangle and to the extensions of the other two sides. Understanding the exradii is imperative as they represent the radii of circles that are associated with the triangle's exterior.

Each exradius is associated with a particular vertex and is inversely proportional to the distance from that vertex to the point where the excircle touches the opposite side (or its extension). The formula to calculate an exradius in a right-angled triangle can be derived using the semiperimeter and the lengths of the sides. A fascinating characteristic of exradii is shown in the equations \( r_1 + r_2 + r_3 = r + R \) and \( r_3 = r + c \). These highlight how the sum of exradii relates to the inradius and the circumradius of the triangle.

To simplify these concepts for students, it's beneficial to visualise the locations of the excircles and relate them with the sides of the triangle, making it more evident how the exradii correspond to the other triangle parameters.