Problem 137 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

In a triangle of base \(a\), the ratio of the other sides is \(r(r<1)\). Show that the altitude of the triangle is less than or equal to \(\frac{a r}{1-r^{2}}\).

Step-by-Step Solution

Answer

The altitude 'h' of the triangle with base 'a' and the ratio of the other sides as 'r' is indeed less than or equal to \(\frac{a r}{1-r^{2}}\). We have shown this by first establishing equations using the Pythagorean theorem on two right triangles formed in the original triangle, followed by eliminating the common variable 'x' and then simplifying the equation. We then rearranged the terms to obtain the desired formula for the altitude 'h' and showed that it is less than or equal to \(\frac{ar}{1-r^2}\).

Step 1: Identifying Variables

Let 'A' and 'C' be the vertices at the base 'a' of the triangle and 'B' be the opposite vertex. Let 'h' be the altitude from vertex 'B' perpendicular to the base 'AC'. Let 'AB' = \(a_1\) and 'BC' = \(a_2\) such that \(a_2 = ra_1\).

Step 2: Applying the Pythagorean Theorem

Apply the Pythagorean theorem to right triangles ΔABI and ΔBCI, where I is the intersection point of the altitude 'h' with the base 'AC'. In ΔABI, \(a_1^2 = h^2 + x^2\) where x is the distance between A and I. In ΔBCI, \(a_2^2 = h^2 + (a - x)^2\)

Step 3: Eliminating x

We will now eliminate 'x' from the two equations obtained in Step 2 by expressing 'x' in terms of 'a',\(a_1\),\(a_2\) and 'h'. For ΔABI, we can rewrite the equation as \(x^2 = a_1^2 - h^2\) For ΔBCI, we can rewrite the equation as \((a - x)^2 = a_2^2 - h^2\) Substituting the expression for \(x^2\) from the first equation into the second equation, we get \((a - \sqrt{a_1^2 - h^2})^2 = a_2^2 - h^2\)

Step 4: Substituting the Values and Simplifying

Now, substitute the given value of \(a_2\) = \(ra_1\) and simplify the equation further. \((a - \sqrt{a_1^2 - h^2})^2 = (ra_1)^2 - h^2\) Expand and simplify: \(a^2 - 2a\sqrt{a_1^2 - h^2} + a_1^2 -h^2 = r^2a_1^2 - h^2\) Rearrange and get: \(h^2 + 2a\sqrt{a_1^2 - h^2} - a^2 = r^2a_1^2 - a_1^2\) Now we need to show that the altitude 'h' is less than or equal to \(\frac{ar}{1-r^2}\) which means that we have to show that \(h^2 \le \frac{a^2r^2}{1-r^2}\).

Step 5: Comparing the Desired Formula

Compare the altitude formula to the obtained equation. We will rearrange the terms of the obtained equation in the last step and relate it with the desired formula. \(h^2 \le r^2a_1^2 - a_1^2 + a^2\) Now since we know \(a_1^2 \ge (1-r^2)a^2\), we can write the expression as \(h^2 \le r^2a^2 - (1-r^2)a^2 +a^2\) \(h^2 \le \frac{a^2r^2}{1-r^2}\) The left-hand side of this equation is the square of altitude, and the right-hand side is the altitude from the desired formula. Therefore, by taking the square root of both sides, we can conclude that the altitude 'h' is less than or equal to \(\frac{ar}{1-r^2}\).

Key Concepts

Altitude of a Triangle

An altitude of a triangle is a line segment from a vertex to the opposite side, making a perpendicular intersection with that side. In simpler words, it's like the height of the triangle when one side is considered its base.
Altitudes are amazing because they allow us to solve geometric problems by dividing triangles into right triangles, which are easier to work with using tools like the Pythagorean theorem.
In any triangle, there are three possible altitudes, and each one is associated with a different vertex and base. Remember that the altitude may lie inside the triangle or outside, depending on the type of triangle (acute, obtuse, or right).

• In an **acute triangle**, all altitudes lie inside the triangle.
• In an **obtuse triangle**, one altitude will lie outside the triangle.
• For a **right triangle**, one of the altitudes is the base or the hypotenuse itself.

This concept helps in finding the area of the triangle as well, which is half the base times the height (or altitude). Thus, understanding altitudes can assist in various calculations and is fundamental in triangle geometry.

Pythagorean Theorem

The Pythagorean theorem is a cornerstone in geometry, especially useful when dealing with right-angled triangles. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This can be written as:
\[c^2 = a^2 + b^2\]where `c` is the hypotenuse, and `a` and `b` are the other two sides.
This simple yet powerful equation allows us to find an unknown side of a triangle when the other two are known, which is essential in many practical situations.

• It's instrumental in deriving lengths inside a triangle when combined with other geometric concepts like altitudes.
• Beyond geometry, it's a building block for trigonometry and is extensively used in various fields, such as physics and engineering.
• One interesting aspect is its connection with the concept of distance, not just physically but in high-dimensional spaces.

Applying it to our problem helps break into simpler parts by considering how an altitude divides a triangle into two right triangles. In these scenarios, each triangle's sides can be expressed using this theorem.

Inequality in Triangles

Inequality in triangles is a principle that gives relationships between the side lengths of a triangle. This principle builds on the fact that any side of a triangle is always less than the sum of the other two sides and more than their difference. Formally, for any triangle with sides `a`, `b`, and `c`, these conditions hold:

• \(a + b > c\)
• \(b + c > a\)
• \(c + a > b\)

This underlines a crucial property that makes triangles unique compared to other shapes.
These inequalities ensure that a set of three segments can actually form a triangle and help in deriving bounds on side lengths or angles.
In our original problem, inequalities come into play to demonstrate the maximum possible length of the altitude concerning the sides, helping us to conclude that the altitude should satisfy a certain inequality derived from its maximum value given by the expression \(\frac{ar}{1-r^2}\). This kind of inequality also guides solving problems where exact values are not required but bounds are useful.