Problem 63 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

On the sides \(\mathrm{AB}, \mathrm{BC}, \mathrm{CA}\) of a \(\triangle \mathrm{ABC}, 3,4,5\) distinct points (excluding vertices \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) ) are respectively chosen. The number of triangles that can be constructed using these chosen points as vertices are: (a) 210 (b) 205 (c) 215 (d) 220

Step-by-Step Solution

Answer

The number of triangles is 205.

Step 1: Understanding the Problem

We need to find the number of triangles that can be formed using distinct points chosen on the sides of a triangle \(\triangle ABC\). We have 3 points on side \(\overline{AB}\), 4 points on side \(\overline{BC}\), and 5 points on side \(\overline{CA}\), excluding the vertices \(A\), \(B\), and \(C\).

Step 2: Calculate Total Points and Combinations

Total points available to pick from are \(3 + 4 + 5 = 12\). We need to choose any 3 points from these 12 points to form a triangle. The number of combinations is calculated using \(\binom{12}{3}\).

Step 3: Calculate \(\binom{12}{3}\)

Use the combination formula \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\), where \(n = 12\) and \(r = 3\). Thus:\[\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220\]

Step 4: Shortlist Valid Combinations

Substitute any points forming a straight line (collinear). Points chosen have to be from different sides. We need to subtract the invalid triangles: - Points only on \(\overline{AB}\): \(\binom{3}{3} = 1\)- Points only on \(\overline{BC}\): \(\binom{4}{3} = 4\)- Points only on \(\overline{CA}\): \(\binom{5}{3} = 10\)The total invalid triangles = 1 + 4 + 10 = 15.

Step 5: Calculate Valid Triangles

Finally, subtract the number of invalid triangles from the initial combination count: \[220 - 15 = 205\]

Step 6: Conclusion

The number of triangles that can be formed using these points, without forming a straight line, is 205.

Key Concepts

Geometry

Geometry is the mathematical study focused on properties, measurements, and relationships of points, lines, angles, surfaces, and solids. In the context of this exercise, we are examining a geometric figure, the triangle, and exploring how different points on its sides can form new triangles. Geometry is essential in understanding shapes and structures, allowing us to visualize spatial concepts.

It's important to know that geometry deals with different types of shapes, each with its own properties. In our exercise, the figure in question is a triangle. Triangles are polygonal shapes with three edges and three vertices, and they are the simplest of all polygons. For any triangle, the sum of its internal angles is 180 degrees. Additionally, the sides of a triangle can vary in length, creating different types of triangles, such as isosceles, equilateral, or scalene.

Triangles

Triangles are a fundamental shape in geometry that consist of three sides and three angles. They are frequently used in various mathematical problems due to their straightforward structure.

In our exercise, triangle \( \triangle ABC \) has sides referred to as \( \overline{AB} \), \( \overline{BC} \), and \( \overline{CA} \), with extra points placed along these lines excluding the original vertices. This setup invites us to think about the combinations of these points to form new triangles. When discussing triangles, it's useful to consider not just their sides and angles, but also their orientation and combination possibilities. Depending on how points align, we can determine whether a new shape or triangle is formed.

Combinations

Combinations in mathematics refer to the selection of items from a larger set without considering the order of selection. In our problem, we used combinations to determine how many triangles can be created using chosen points.

When determining combinations, we use the binomial coefficient formula: \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of items, and \( r \) is the number of items to choose. In our case, from 12 possible points, we choose 3 to form a triangle, which gives us \( \binom{12}{3} \). After calculating, it initially counts 220 possible triangles. However, through deeper understanding, we subtract those that do not fulfill our triangle criteria, focusing on ensuring each triangle's vertices are chosen from different sides.

Collinearity

Collinearity is a property where three or more points lie on the same straight line. In our context, collinear points cannot form a triangle as a triangle's vertices must be non-linear. If they are on the same line, the so-called 'triangle' collapses into a line, not a true triangle.

Understanding collinearity is essential because, in our solution, we needed to subtract combinations of points where all were from a single side. For instance, the points selected from any one side alone (like \( \overline{AB} \), \( \overline{BC} \), or \( \overline{CA} \)) are collinear. We calculated such invalid setups—1 from \( \overline{AB} \), 4 from \( \overline{BC} \), and 10 from \( \overline{CA} \), totaling 15 collinear cases. These invalid setups are subtracted from the total possible triangles, giving us the valid count.