Math Problem Statement

Consider a square ABCD of side length 16. Let E, F be points on CD such that CE = EF = FD. Let the line BF and AE meet in M. The area of triangle MAB is

Solution

To solve the problem, let's first visualize the square ABCDABCD with side length 1616, and place it on a coordinate plane:

  • Assume point A=(0,16)A = (0, 16), B=(16,16)B = (16, 16), C=(16,0)C = (16, 0), and D=(0,0)D = (0, 0).

Points EE and FF divide side CDCD into three equal segments:

  1. Since CD=16CD = 16, each segment CE=EF=FD=1635.33CE = EF = FD = \frac{16}{3} \approx 5.33.

  2. The coordinates for EE and FF are:

    • E=(16163,0)=(323,0)E = \left(16 - \frac{16}{3}, 0\right) = \left(\frac{32}{3}, 0\right)
    • F=(162×163,0)=(163,0)F = \left(16 - 2 \times \frac{16}{3}, 0\right) = \left(\frac{16}{3}, 0\right)

Step 1: Find Equations of Lines AE and BF

Line AEAE:

  1. Point A=(0,16)A = (0, 16) and E=(323,0)E = \left(\frac{32}{3}, 0\right).
  2. Slope of line AEAE is: slope of AE=0163230=16323=16332=32\text{slope of } AE = \frac{0 - 16}{\frac{32}{3} - 0} = -\frac{16}{\frac{32}{3}} = -\frac{16 \cdot 3}{32} = -\frac{3}{2}
  3. The equation of line AEAE in point-slope form is: y16=32x    y=32x+16y - 16 = -\frac{3}{2}x \implies y = -\frac{3}{2}x + 16

Line BFBF:

  1. Point B=(16,16)B = (16, 16) and F=(163,0)F = \left(\frac{16}{3}, 0\right).
  2. Slope of line BFBF is: slope of BF=01616316=1616316=1616483=16332=4832=32\text{slope of } BF = \frac{0 - 16}{\frac{16}{3} - 16} = \frac{-16}{\frac{16}{3} - 16} = \frac{-16}{\frac{16 - 48}{3}} = \frac{-16 \cdot 3}{-32} = \frac{48}{32} = \frac{3}{2}
  3. The equation of line BFBF in point-slope form is: y16=32(x16)    y=32x8y - 16 = \frac{3}{2}(x - 16) \implies y = \frac{3}{2}x - 8

Step 2: Find Coordinates of Point M

To find the intersection MM of lines AEAE and BFBF, set their equations equal:

32x+16=32x8-\frac{3}{2}x + 16 = \frac{3}{2}x - 8

Combine like terms:

16+8=32x+32x    24=3x    x=816 + 8 = \frac{3}{2}x + \frac{3}{2}x \implies 24 = 3x \implies x = 8

Substitute x=8x = 8 back into either line equation to find yy:

y=328+16=12+16=4y = -\frac{3}{2} \cdot 8 + 16 = -12 + 16 = 4

Thus, point M=(8,4)M = (8, 4).

Step 3: Calculate the Area of Triangle MAB

The vertices of triangle MABMAB are M=(8,4)M = (8, 4), A=(0,16)A = (0, 16), and B=(16,16)B = (16, 16).

The area of a triangle given by coordinates (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), (x3,y3)(x_3, y_3) is:

Area=12x1(y2y3)+x2(y3y1)+x3(y1y2)\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|

Substitute the coordinates:

Area=128(1616)+0(164)+16(416)\text{Area} = \frac{1}{2} \left| 8(16 - 16) + 0(16 - 4) + 16(4 - 16) \right|

Simplify:

Area=120+0+16(12)=12192=1922=96\text{Area} = \frac{1}{2} \left| 0 + 0 + 16(-12) \right| = \frac{1}{2} \left| -192 \right| = \frac{192}{2} = 96

Conclusion

The area of triangle MABMAB is 96\boxed{96}.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Algebra
Geometry
Triangles

Formulas

Slope formula: m = (y2 - y1) / (x2 - x1)
Equation of a line: y - y1 = m(x - x1)
Area of a triangle: (1/2) * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Theorems

Triangle area formula using coordinates

Suitable Grade Level

Grades 10-12