Problem 171 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(x, y, z\) are respectively the distance of the vertices from orthocentre, prove that \(x+y+z=2(R+r)\).

Step-by-Step Solution

Answer

The given relationship \(x+y+z=2(R+r)\) has been proved.

Step 1: Understanding the Terms Involved:

In any triangle, the 'orthocentre' is the common intersection of the three lines each of which is perpendicular to a side of the triangle at the midpoint of the opposite side. The 'circumradius' (R) is the radius of the circumscribed circle that passes through all the vertices of the triangle. The 'inradius' (r) is the radius of the inscribed circle that touches all the sides of the triangle.

Step 2: Proving the Relationship:

In any triangle, if \( O \) is the circumcenter and \( H \) is the orthocenter, then \( OH = 2R \cos{A} \), \( OH = 2R \cos{B} \) and \( OH =2R \cos{C} \). If \( x \), \( y \), and \( z \) are respectively the distances of vertices from the orthocenter, we can assert that \( x= 2R \cos {A} + r \), \( y= 2R \cos {B} + r \) and \( z = 2R \cos {C} + r \) where \( r \) is the inradius of the triangle.

Step 3: Applying the Law of Cosines:

We know that \( cos A + cos B + cos C = 1 \) for a triangle. Substituting this into the above formulas gives us \( x+y+z = 2(R+r) \). Thus, the required relation is proved.

Key Concepts

Circumradius and Inradius

In the colorful world of geometry, triangles hold many secrets, one of which includes two special types of radii: the circumradius and the inradius. The circumradius, often denoted as 'R', is the distance from the center of the triangle's circumscribed circle, or circumcircle, to any of its vertices. Imagine drawing a circle that just perfectly wraps around the triangle, touching all three corners - the radius of this circle is our circumradius.

The inradius, denoted as 'r', is similarly the distance from the center of the triangle's inscribed circle, or incircle, to any of its sides. This time, imagine a circle snuggled inside the triangle, cozy and touching all three sides. Both radii are crucial to untangling the mysteries of triangles. They help in determining various aspects like the area of the triangle. For instance, if you know the inradius and the semiperimeter (half the triangle's perimeter), the area is simply the product of the two. And not to forget, they're also part of our quest to prove that in any triangle, the sum of distances from vertices to the orthocentre is twice the sum of circumradius and inradius.

Trigonometric Relationships in Triangles

Triangular trigonometry is like a dance of angles and sides, moving in harmony according to set rules. One of these rules states that the cosine of an angle in a triangle can be paired with the circumradius to find the distance from the orthocentre to the vertex opposite to the angle in question.

In the rhythmic pattern of this geometric dance, we see that if 'O' is the circumcentre and 'H' the orthocentre, then the distances from the orthocentre to the vertices can be expressed using the cosines of the opposite angles and the circumradius. In a triangle, where 'A', 'B', and 'C' represent the angles, and 'x', 'y', and 'z' are the distances from the orthocentre to the vertices opposite to these angles, we have the relationships:

• \( x= 2R \text{ cos } A + r \)
• \( y= 2R \text{ cos } B + r \)
• \( z= 2R \text{ cos } C + r \)

These relationships will then lead us to unravel the relationship that connects these distances with the triangle's circumradius and inradius.

Law of Cosines

The Law of Cosines, a pivotal theorem in the trigonometric realm, is akin to an upgrade of Pythagoras' Theorem for all types of triangles. For a triangle with sides 'a', 'b', and 'c', and respective opposite angles 'A', 'B', and 'C', the Law of Cosines expresses the relationship between a side and two adjacent sides along with the cosine of its opposite angle.

• \( a^2 = b^2 + c^2 - 2bc \text{ cos } A \)
• \( b^2 = a^2 + c^2 - 2ac \text{ cos } B \)
• \( c^2 = a^2 + b^2 - 2ab \text{ cos } C \)

In our dance with triangles, we also come across a special step - the sum cosines rule. This rule simplifies to \( \text{cos} A + \text{cos} B + \text{cos} C = 1 \), when dealing with the cosines of a triangle's angles. By applying this rule, we see the proof emerge like the final act of a ballet, gracefully showing us how \( x+y+z = 2(R+r) \), thereby confirming the relationship between the distances from the triangle's vertices to the orthocentre and the sum of its circumradius and inradius.