A geometry problem by Lil Phil

Geometry Level 3

The area of the shaded rectangle is:

between 1/4 and 5/16 more than 1/2 between 7/16 and 1/2 between 5/16 and 3/8 between 3/8 and 7/16

Lil Phil by Lil Phil , 21, Antarctica

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3 solutions

David Vreken
Jul 3, 2019

Label the outer rectangle ABCD and the inner shaded rectangle EFHG as follows:

Using angle sums of triangles and straight lines are 180 ° 180° , it can be determined that the four white right triangles are similar by AA similarity.

From △ A E D ∼ △ B F E \triangle AED \sim \triangle BFE , B F B E = A E A D \frac{BF}{BE} = \frac{AE}{AD} or B F 1 2 = 1 2 1 \frac{BF}{\frac{1}{2}} = \frac{\frac{1}{2}}{1} , which solves to B F = 1 4 BF = \frac{1}{4} .

By Pythagorean's Theorem on △ B F E \triangle BFE , E F = 5 4 EF = \frac{\sqrt{5}}{4} .

Since B F + F C = 1 BF + FC = 1 and B F = 1 4 BF = \frac{1}{4} , F C = 3 4 FC = \frac{3}{4} .

From △ B F E ∼ △ C G F \triangle BFE \sim \triangle CGF , F G F C = E F E B \frac{FG}{FC} = \frac{EF}{EB} or F G 3 4 = 5 4 1 2 \frac{FG}{\frac{3}{4}} = \frac{\frac{\sqrt{5}}{4}}{\frac{1}{2}} , which solves to F G = 3 5 8 FG = \frac{3\sqrt{5}}{8} .

The area of the shaded rectangle is then A E F G H = E F ⋅ F G = 5 4 ⋅ 3 5 8 = 15 32 A_{EFGH} = EF \cdot FG = \frac{\sqrt{5}}{4} \cdot \frac{3\sqrt{5}}{8} = \frac{15}{32} , which is between 7 16 \frac{7}{16} and 1 2 \frac{1}{2} .

4 Helpful 1 Interesting 3 Brilliant 0 Confused

I did not think of figuring out if the triangles were similar, I ended up using coordinate geometry which raised up a lot of equations. Very nice observation!

José Alejandro - 1 year, 11 months ago
Steven Chase
Jul 3, 2019 

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import math

##############################
# Point O

Ox = 0.0
Oy = 0.0
##############################

# Point S

Sx = 0.5
Sy = 1.0
##############################

# Unit Vector u
vx = Ox - Sx
vy = Oy - Sy

v = math.hypot(vx,vy)

ux = vx/v
uy = vy/v
##############################

# Unit Vector u2
u2x = -uy
u2y = ux

##############################

# Point T coordinates and shorter side length alpha

# Sx + alpha*u2x = 1
alpha = (1.0-Sx)/u2x

Tx = Sx + alpha*u2x
Ty = Sy + alpha*u2y

##############################

# Longer side length beta

# Ty + beta*uy = 0
beta = -Ty/uy

##############################

# Area calculation

Area = alpha * beta

##############################

# Compare area to ranges
print Area
print ""

print (7.0/16.0),(1.0/2.0)
print (1.0/4.0),(5.0/16.0)
print (5.0/16.0),(3.0/8.0)
print (3.0/8.0),(7.0/16.0)

#0.46875

#0.4375 0.5
#0.25 0.3125
#0.3125 0.375
#0.375 0.4375

1 Helpful 0 Interesting 1 Brilliant 0 Confused

Length of the shaded rectangle is 3 √ 5 8 \dfrac{3√5}{8} and breadth is √ 5 4 \dfrac{√5}{4} . Therefore it's area is 15 32 \dfrac{15}{32} , which lies between 7 16 \dfrac{7}{16} and 1 2 \dfrac{1}{2}

1 Helpful 0 Interesting 1 Brilliant 1 Confused

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