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Problem of the Week
Problem E and Solutions
Building In
C
Problem
In the diagram, 4ABC is right-angled and AB = 2BC. D lies
on AB, E lies on AC, and F lies on BC such that DEF B is
a square. Determine the ratio of the area of square DEF B to
4ABC.
A
B
Solution
Draw in square DEF B. Let BF = a and F C = b.
Since DEF B is a square, DE = EF = DB = BF = a.
Also, AB = 2BC = 2(BF + F C) = 2(a + b) = 2a + 2b.
Since AB = AD + DB, we have 2a + 2b = AD + a, or
AD = a + 2b.
In Solution 1, we will solve the problem using similar triangles.
In Solution 2, we will place the diagram on the xy-plane and
solve the problem using analytic geometry.
C
E
a
a
A
a + 2b
D
Solution 1
Consider 4EF C and 4ABC. We will first show that 4EF C ∼ 4ABC.
Since DEF B is a square, ∠EF B = 90◦ , and so ∠EF C = 180◦ − ∠EF B = 180◦ − 90◦ = 90◦ .
Therefore, ∠EF C = ∠ABC. Also, ∠ECF = ∠ACB since they represent the same angle.
Since the angles in a triangle add to 180◦ , we must also have ∠CEF = ∠CAB.
So 4EF C ∼ 4ABC, by Angle−Angle−Angle Triangle Similarity.
Since 4EF C ∼ 4ABC, corresponding side lengths are in the same ratio. In particular,
CF
BC
=
EF
AB
CF
BC
=
EF
2BC
b
1
=
a
2
a = 2b
Since AB = 2a + 2b and a = 2b, AB = 2(2b) + 2b = 6b.
Since BC = a + b and a = 2b, BC = 2b + b = 3b.
The area of 4ABC is 21 (AB × BC) = 12 (6b × 3b) = 9b2 .
The area of square DEF B is a × a = a2 = (2b)2 = 4b2 .
The ratio of the area of inscribed square DEF B to 4ABC is 4b2 : 9b2 = 4 : 9, since b > 0.
b
F
a
a
B
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Solution 2
We proceed by placing the triangle on the xy−plane with A at (0, 0) and AB along the positive
x−axis. The coordinates of E are (a + 2b, a) and the coordinates of C are (2a + 2b, a + b).
y
C (2a + 2b, a + b)
E (a + 2b, a)
b
F
a
A (0, 0)
a + 2b
D
a
x
B
Let’s determine the equation of the line through A, E, C.
Since this line passes through (0, 0), we know it has y-intercept 0.
a−0
Since it passes through (0, 0) and (a + 2b, a), the line has slope = a+2b−0
=
a
Therefore, the equation of the line through A, E, C is y = a+2b x.
a
.
a+2b
Since C(2a
+ 2b, a + b) lies on this line, substituting x = 2a + 2b and y = a + b into
a
y = a+2b
x, we have
a+b =
(a + b)(a + 2b)
a2 + 3ab + 2b2
2b2 + ab − a2
(2b − a)(b + a)
a = 2b
=
=
=
=
or
a
(2a + 2b)
a + 2b
a(2a + 2b)
2a2 + 2ab
0
0
a = −b
(1)
But since a, b > 0, a = −b is inadmissible and we must have a = 2b.
Since AB = 2a + 2b and a = 2b, AB = 2(2b) + 2b = 6b.
Since BC = a + b and a = 2b, BC = 2b + b = 3b.
The area of 4ABC is 21 (AB × BC) = 12 (6b × 3b) = 9b2 .
The area of square DEF B is a × a = a2 = (2b)2 = 4b2 .
The ratio of the area of inscribed square DEF B to 4ABC is 4b2 : 9b2 = 4 : 9, since b > 0.
Note that from (1), (a + b)(a + 2b) = a(2a + 2b), we obtain (a + b)(a + 2b) = 2a(a + b) by
factoring the right side of the equation.
Since a, b > 0, a + b > 0 and we could have divided out the common factor leaving a + 2b = 2a
which simplifies to a = 2b. Then the trinomial factoring to determine a = 2b would not have
been necessary.