CAT 2024 Slot 3 QA Question 11

Okay, so I have this geometry problem from the CAT exam. Let me try to understand it step by step. The question is about a triangle ABC with midpoints M, N, and P on sides AB, BC, and AC respectively. The medians from A, B, and C intersect the line segments MP, MN, and NP at points X, Y, and Z. The area of triangle ABC is given as 1440 sq \text{ cm}, and We find the area of triangle XYZ.

First, I need to visualize the triangle and the points mentioned. Let me sketch a rough triangle ABC. Let's label the vertices A, B, and C. Now, the midpoints: M is the midpoint of AB, N is the midpoint of BC, and P is the midpoint of AC. So, connecting these midpoints, MP, MN, and NP are the midlines of the triangle.

the midline of a triangle is parallel to the third side and half its length. So, MP is parallel to BC and half its length, MN is parallel to AC and half its length, and NP is parallel to AB and half its length.

Now, the medians from A, B, and C intersect these midlines at X, Y, and Z respectively. Medians are the lines from a vertex to the midpoint of the opposite side. So, the median from A goes to N (midpoint of BC), the median from B goes to P (midpoint of AC), and the median from C goes to M (midpoint of AB).

I think it might help to assign coordinates to the triangle to make this more concrete. Let me place triangle ABC in a coordinate system. Let's assume point A is at (0, 0), point B is at (2b, 0), and point C is at (2c, 2d). Choosing these coordinates with even numbers might make the midpoints have integer coordinates, which could simplify calculations.

So, midpoint M of AB would be at ((0 + 2b)/2, (0 + 0)/2) = (b, 0). Midpoint N of BC would be at ((2b + 2c)/2, (0 + 2d)/2) = (b + c, d). Midpoint P of AC would be at ((0 + 2c)/2, (0 + 2d)/2) = (c, d).

Now, the midlines: MP connects M(b, 0) to P(c, d). MN connects M(b, 0) to N(b + c, d). NP connects N(b + c, d) to P(c, d).

Next, We find the equations of these midlines and the equations of the medians, then find their intersection points X, Y, Z.

First, let's find the equations of the medians.

Median from A: This is the line from A(0, 0) to N(b + c, d). The slope of this median is (d - 0)/(b + c - 0) = d/(b + c). So, the equation is y = (d/(b + c))x.

Median from B: This is the line from B(2b, 0) to P(c, d). The slope is (d - 0)/(c - 2b) = d/(c - 2b). The equation can be written using point-slope form: y - 0 = [d/(c - 2b)](x - 2b), so y = [d/(c - 2b)](x - 2b).

Median from C: This is the line from C(2c, 2d) to M(b, 0). The slope is (0 - 2d)/(b - 2c) = (-2d)/(b - 2c) = (2d)/(2c - b). The equation using point-slope form: y - 2d = [2d/(2c - b)](x - 2c).

Now, let's find the equations of the midlines.

Midline MP: Connects M(b, 0) to P(c, d). The slope is (d - 0)/(c - b) = d/(c - b). The equation is y - 0 = [d/(c - b)](x - b), so y = [d/(c - b)](x - b).

Midline MN: Connects M(b, 0) to N(b + c, d). The slope is (d - 0)/( (b + c) - b ) = d/c. The equation is y - 0 = (d/c)(x - b), so y = (d/c)(x - b).

Midline NP: Connects N(b + c, d) to P(c, d). Since both points have the same y-coordinate, d, this is a horizontal line. So, the equation is y = d.

Now, let's find the intersection points X, Y, Z.

Point X is the intersection of the median from A (y = (d/(b + c))x) and midline MP (y = [d/(c - b)](x - b)).

Set the two equations equal:

(d/(b + c))x = [d/(c - b)](x - b)

We can cancel d from both sides (assuming d ≠ 0, which it isn't since the area is 1440):

(1/(b + c))x = (1/(c - b))(x - b)

Multiply both sides by (b + c)(c - b) to eliminate denominators:

(c - b)x = (b + c)(x - b)

Expand the right side:

(c - b)x = (b + c)x - (b + c)b

Bring all terms to the left:

(c - b)x - (b + c)x + (b + c)b = 0

Factor x:

[(c - b) - (b + c)]x + (b + c)b = 0

Simplify the coefficient of x:

(c - b - b - c) = (-2b)

So:

-2b x + (b + c)b = 0

Solve for x:

-2b x = - (b + c)b

Divide both sides by -2b (assuming b ≠ 0):

x = [(b + c)b]/(2b) = (b + c)/2

Now, plug x back into the median from A's equation to find y:

y = (d/(b + c)) * ( (b + c)/2 ) = d/2

So, point X is at ( (b + c)/2 , d/2 )

Similarly, let's find point Y, the intersection of the median from B and midline MN.

Median from B: y = [d/(c - 2b)](x - 2b)

Midline MN: y = (d/c)(x - b)

Set them equal:

[d/(c -