Area of ABC?

Geometry Level 3

The figure consists of 25 equilateral triangles, each of area 1 c m 2 cm^2 . What is the area of △ A B C = ? \triangle ABC=?

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Hana Wehbi by Hana Wehbi , 51, USA

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4 solutions

David Vreken
May 16, 2019

The whole diagram can be sheared to a rectangular grid while preserving area:

The triangle goes through B = 4 B = 4 boundary points and has I = 4 I = 4 interior points. If the rectangles were unit squares, then by Pick's Theorem the area would be A = I + 1 2 B − 1 = 4 + 1 2 ⋅ 4 − 1 = 5 A = I + \frac{1}{2}B - 1 = 4 + \frac{1}{2} \cdot 4 - 1 = 5 .

However, since each rectangle is made up of 2 2 unit triangles for an area of 2 2 instead of 1 1 , the area is doubled, and so A = 2 I + B − 2 = 2 ⋅ 4 + 4 − 2 = 10 A = 2I + B - 2 = 2 \cdot 4 + 4 - 2 = \boxed{10} .

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Good use of Pick's Transformation for solving this problem. https://en.wikipedia.org/wiki/Schwarz lemma#Schwarz%E2%80%93Pick theorem (*Holomorphic Indices)

nibedan mukherjee - 2 years ago
Nibedan Mukherjee
May 14, 2019

*N.B. Flaw " Area of (BRS) " correction " Area of (BRC). Alternate solution;

2 Helpful 0 Interesting 2 Brilliant 0 Confused
Hana Wehbi
May 15, 2019

From the figure note that we have 18 yellow triangles. If x x is the area of △ A B C \triangle ABC then 2 ( x − 1 ) = 18 ⟹ x = 10 2(x-1)=18\implies x=\boxed{10}

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Steven Chase
May 14, 2019 
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import math

A = 1.0 # small triangle area

a = math.sqrt(4.0*A/math.sqrt(3.0)) # small triangle side length

Cx = a    # Point C coordinates
Cy = 0.0

ang1 = math.pi/3.0      # angle from bottom left to point A
ang2 = 2.0*math.pi/3.0  # angle from bottom right to point B  

u1x = math.cos(ang1)  # unit vector from bottom left to point A
u1y = math.sin(ang1)

u2x = math.cos(ang2)  # unit vector from bottom right to point B
u2y = math.sin(ang2)  

Ax = 4.0*a*u1x   # coordinates of point A
Ay = 4.0*a*u1y

Bx = 5.0*a + 2.0*a*u2x  # coordinates of point B
By = 0.0 + 2.0*a*u2y

v1x = Ax - Cx  # vector from point C to point A
v1y = Ay - Cy

v2x = Bx - Cx  # vector from point C to point B
v2y = By - Cy

cross = v1x*v2y - v1y*v2x  # vector cross product

AreaABC = 0.5*math.fabs(cross)   # area of ABC is half cross product magnitude

print AreaABC  # AreaABC = 10

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