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Constructing a right triangle (Posted on 2008-12-07) Difficulty: 2 of 5
Construct a right triangle of area 8 out of three other right triangles with areas 1, 3 and 4.

See The Solution Submitted by pcbouhid    
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Solution Generalized soltution | Comment 4 of 6 |
B
+_
|\\_
| \ \_
|  \  \_
|   \   \_E
|    \   /\_
|     \ /   \
+------+-----+
C      D     A
The large triangle ABC (with right angle C) is divided into triangles BCD (with right angle C), BDE (with right angle E), and ADE (with right angle E).

Let the area of BCD be a, let the area of BDE be b, and let the area of ADE be c.

Triangles BCD and ABD have the same altitude BC, so their areas are proportinate to the bases: a/(b+c) = CD/DA. Let x be a variable such that CD=a*x and DA=(b+c)*x.

Triangles BDE and ADE have the same altitude DE, so their areas are proportinate to the bases: b/c = BE/EA. Let y be a variable such that BE=b*y and EA=c*y.

Let p be the length of BD, let q be the length of DE, and let z be the length of BC.

Using the area formula and the Pythagorean theorem, six equations can be formed:
z^2 + (a*x)^2 = p^2
q^2 + (b*y)^2 = p^2
q^2 + (c*y)^2 = [(b+c)*x]^2
z^2 + [(a+b+c)*x]^2 = [(b+c)*y]^2
x*z = 2
q*y = 2

Combining the first three equations to remove p^2 and q^2 yields:
z^2 = b^2*y^2 - c^2*y^2 + b^2*x^2 + 2*b*c*x^2 + c^2*x^2 - a^2*x^2

Combining the two equations for z^2 gives:
2*b*c*y^2 + 2*c^2*y^2 = 2*b^2*x^2 + 2*a*b*x^2 + 2*a*c*x^2 + 4*b*c*x^2 + 2*c^2*x^2
After dividing by 2*C*(b+c):
c*y^2 = (a+b+c)*x^2

Substituting back into an equation for z^2 yields:
z^2 = (b^2 + b*c - a*c)*y^2
Note that this last equation places a restriction in the values of a, b, and c, namely b^2 + b*c - a*c > 0

Combining the last two equations with x*z=2 from earlier expresses y in terms of a, b, and c:
y^4 = [4*(a+b+c)]/[c*(b^2+b*c-a*c)] To simplify later expressions, let s=a+b+c and d=b^2+b*c-a*c.
Then y^4 = (4*s)/(c*d)

Substituting this expression for y into the two prior equations for z and x
x^4 = (4*c)/(s*d)
z^4 = (4*s*d)/c

Substituting the previous expressions into equations for p and q:
p^4 = [4*(b+c)^2*(d+a*b)^2]/[c*s*d]
q^4 = (4*c*d)/s

Then expressions for all the sides can be created with the following summary:
area(BCD) = a
area(BDE) = b
area(ADE) = c
area(ABC) = s = a+b+c
d = b^2+b*c-a*c > 0
AD = (b+c) * 4throot[(4*c)/(s*d)]
DC = a * 4throot[(4*c)/(s*d)]
AC = s * 4throot[(4*c)/(s*d)]
BC = 4throot[(4*s*d)/c]
BE = b * 4throot[(4*s)/(c*d)]
EA = c * 4throot[(4*s)/(c*d)]
BA = (b+c) * 4throot[(4*s)/(c*d)]
BD = 4throot[4*(b+c)^2*(d+a*b)^2/(c*s*d)]
DE = 4throot[(4*c*d)/s]


Four of the six possible ways of assigning the values 1,3,4 to a,b,c yield positive values for d:
a=1, b=3, c=4: d=17
AB = 7 * 4throot(8/17), AC = 8 * 4throot(2/17), BC = 4throot(136)

a=1, b=4, c=3: d=25
AB = 14 * 4throot(2/75), AC = 4 * 4throot(24/25), BC = 2 * 4throot(50/3)

a=3, b=4, c=1: d=17
AB = 10 * 4throot(2/17), AC = 4 * 4throot(8/17), BC = 2 * 4throot(34)

a=4, b=3, c=1: d=8
AB = 4 * sqrt(2), AC = 4, BC = 4

a=4, b=1, c=3: d = -8 -> No triangle

a=3, b=1, c=4: d = -7 -> No triangle
  Posted by Brian Smith on 2008-12-08 02:49:52
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