Problem 107 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(p_{1}, p_{2}, p_{3}\) are altitudes of a triangle \(A B C\) from the vertices \(A, B, C\) and \(\Delta\) the area of the triangle, then prove that \(p_{1}^{-2}+p_{2}^{-2}+p_{3}^{-2}=\frac{a^{2}+b^{2}+c^{2}}{4 \Delta^{2}}\).

Step-by-Step Solution

Answer

To prove the equation \(p_{1}^{-2}+p_{2}^{-2}+p_{3}^{-2}=\frac{a^{2}+b^{2}+c^{2}}{4 \Delta^{2}}\), we express the altitudes \(p_1, p_2, p_3\) in terms of the area \(\Delta\) and the sides \(a, b, c\), substitute them in the given equation, and simplify to obtain the desired result: Step 1: Write the area of the triangle using altitudes: \(\Delta = \frac{1}{2} \cdot a \cdot p_1 = \frac{1}{2} \cdot b \cdot p_2 = \frac{1}{2} \cdot c \cdot p_3\). Step 2: Express the altitudes in terms of area and sides: \(p_1 = \frac{2\Delta}{a}\), \(p_2 = \frac{2\Delta}{b}\), \(p_3 = \frac{2\Delta}{c}\). Step 3: Substitute altitudes in the given equation: \(\left(\frac{2\Delta}{a}\right)^{-2}+\left(\frac{2\Delta}{b}\right)^{-2}+\left(\frac{2\Delta}{c}\right)^{-2}=\frac{a^{2}+b^{2}+c^{2}}{4 \Delta^{2}}\). Step 4: Simplify and rearrange the equation to prove the given equation: \(p_{1}^{-2}+p_{2}^{-2}+p_{3}^{-2}=\frac{a^{2}+b^{2}+c^{2}}{4 \Delta^{2}}\).

Step 1: Write the area of the triangle in terms of its sides and altitudes

We are given that the altitudes of the triangle are \(p_1, p_2, p_3\), and the area of the triangle is \(\Delta\). Using the definition of an altitude, we can write the area of the triangle in terms of its sides and altitudes: \[ \Delta = \frac{1}{2} \cdot a \cdot p_1 = \frac{1}{2} \cdot b \cdot p_2 = \frac{1}{2} \cdot c \cdot p_3 \] Where \(a, b, c\) are the lengths of the sides of the triangle opposite to vertices \(A, B, C\), respectively.

Step 2: Express the altitudes in terms of area and sides

From the expressions above, we can write the altitudes in terms of the area and sides: \[ p_1 = \frac{2\Delta}{a} \] \[ p_2 = \frac{2\Delta}{b} \] \[ p_3 = \frac{2\Delta}{c} \]

Step 3: Substitute altitudes in the given equation

Now we need to prove the given equation: \[ p_{1}^{-2}+p_{2}^{-2}+p_{3}^{-2}=\frac{a^{2}+b^{2}+c^{2}}{4 \Delta^{2}} \] Substitute the expressions for the altitudes from Step 2 into the equation: \[ \left(\frac{2\Delta}{a}\right)^{-2}+\left(\frac{2\Delta}{b}\right)^{-2}+\left(\frac{2\Delta}{c}\right)^{-2}=\frac{a^{2}+b^{2}+c^{2}}{4 \Delta^{2}} \]

Step 4: Simplify and rearrange the equation

Simplify the equation by multiplying each term by the corresponding denominator: \[ \frac{a^2}{(2\Delta)^2}+\frac{b^2}{(2\Delta)^2}+\frac{c^2}{(2\Delta)^2}=\frac{a^{2}+b^{2}+c^{2}}{4 \Delta^{2}} \] Subtract both sides by \(\frac{a^{2}+b^{2}+c^{2}}{4 \Delta^{2}}\): \[ \frac{a^2}{(2\Delta)^2}+\frac{b^2}{(2\Delta)^2}+\frac{c^2}{(2\Delta)^2}-\frac{a^{2}+b^{2}+c^{2}}{4 \Delta^{2}}=0 \] Factor out \(\frac{1}{4\Delta^2}\): \[ \frac{1}{4\Delta^2}(a^2+b^2+c^2-(a^2+b^2+c^2))=0 \] Now we have proved the given equation: \[ p_{1}^{-2}+p_{2}^{-2}+p_{3}^{-2}=\frac{a^{2}+b^{2}+c^{2}}{4 \Delta^{2}} \]

Key Concepts

Altitudes in Triangles

Altitudes in a triangle are fascinating geometric lines that play a crucial role in understanding the triangle's dimensions. An altitude is a line segment drawn from a vertex of the triangle, perpendicular to the opposite side, or the line containing the opposite side. It is important to note that every triangle has three altitudes, one from each vertex, intersecting at a single point known as the orthocenter.
An easy way to visualize altitudes is to think of them as 'height' lines for each side when discussing the area of triangles. The length of an altitude gives the shortest distance from a vertex to the line opposite it. This detail makes altitudes distinct from other types of lines you might draw in a triangle.

• Altitudes help in calculating the area of a triangle efficiently using simple formulas.
• They intersect at the orthocenter, a unique point within or outside the triangle, based on its type - acute, obtuse, or right-angled.

Understanding how to find and use altitudes is beneficial not only for solving geometric problems but also for developing spatial reasoning skills.

Area of Triangle

The area of a triangle is a fundamental concept in geometry, representing the amount of space enclosed within the triangle's boundaries. It's usually measured in square units. The area can be calculated using a variety of formulas, but one of the most direct ways involves the base and height, which is where altitudes come into play.
The formula for the area based on a base and its corresponding altitude (height) is:
\[\Delta = \frac{1}{2} \times \text{base} \times \text{height}\]
This formula can be adapted for any of the triangle's sides, as each side can be considered a base with its respective altitude.

• Using altitudes for calculation provides a straightforward method to find the area without the need for complex inputs.
• It clarifies how different geometric aspects of a triangle relate to one another, such as how changing one side or altitude impacts the area.

The use of altitudes in determining the area hinges on a fundamental understanding of perpendicular height, which often simplifies the arithmetic process involved in finding areas in more complex geometric arrangements.

Triangle Sides

The sides of a triangle form the backbone of its geometric structure, defining its shape and attributes. Each side of a triangle can serve as a base when calculating areas with respective altitudes. The triangle side lengths are often denoted by the lowercase letters \(a\), \(b\), and \(c\), corresponding to vertices \(A\), \(B\), and \(C\) respectively.
The relationship between the sides and other components of the triangle, like angles and altitudes, is central in deriving various geometric equations and properties. In the problem at hand, the sides connect with altitudes and the area to form a specific equation that relates the inverse of the squares of altitudes with sides and area.

• Sides play a pivotal role in establishing the perimeters and potential classification of a triangle (scalene, isosceles, or equilateral).
• Sides are essential for understanding the derivation of other triangle properties like the orthocenter, centroid, and inradius.

Recognizing side relations provides the foundation for solving geometric problems, as many properties and theorems in triangles revolve around side lengths and their interactions with other triangle elements.