04-08-2017, 09:17 PM

(04-08-2017 05:19 PM)Han Wrote: [ -> ]For anyone who is curious about the tedious algebra using Heron's formula as shown by Gerson, here is a text output from Maple 17:

Code:

`> restart:`

> eq1:=a^2=b^2+c^2-2*b*c*cosA;

2 2 2

a = -2 b c cosA + b + c

> eq2:=b^2=a^2+c^2-2*a*c*cosB;

2 2 2

b = -2 a c cosB + a + c

> eq3:=c^2=a^2+b^2-2*a*b*cosC;

2 2 2

c = -2 a b cosC + a + b

> s1:=1/2*(a+b+c);

1 1 1

- a + - b + - c

2 2 2

> solve({eq1,eq2,eq3},{cosA,cosB,cosC});

/ 2 2 2 2 2 2 2 2 2\

| a - b - c a - b + c a + b - c |

< cosA = - ------------, cosB = ------------, cosC = ------------ >

| 2 b c 2 a c 2 a b |

\ /

> cosA := -(1/2)*(a^2-b^2-c^2)/(b*c); cosB := (1/2)*(a^2-b^2+c^2)/(a*c); cosC :=

> (1/2)*(a^2+b^2-c^2)/(a*b);

2 2 2

a - b - c

- ------------

2 b c

2 2 2

a - b + c

------------

2 a c

2 2 2

a + b - c

------------

2 a b

> b:=2*r; c:=3*s; a:=4*t;

2 r

3 s

4 t

> eq5:=u^2=s^2+(b-r)^2-2*s*(b-r)*cosA;

2 1 2 1 2 8 2

u = - - s + - r + - t

2 3 3

> eq6:=v^2=t^2+(c-s)^2-2*t*(c-s)*cosB;

2 5 2 5 2 2 2

v = - - t + - s + - r

3 2 3

> eq7:=w^2=r^2+(a-t)^2-2*r*(a-t)*cosC;

2 1 2 2 27 2

w = - - r + 3 t + -- s

2 8

> s2:=1/2*(u+v+w);

1 1 1

- u + - v + - w

2 2 2

>

>

> H1:=expand(s1*(s1-a)*(s1-b)*(s1-c));

2 2 2 2 9 2 2 4 4 81 4

8 t r + 18 t s + - r s - 16 t - r - -- s

2 16

> H2:=expand(s2*(s2-u)*(s2-v)*(s2-w));

1 2 2 1 2 2 1 2 2 1 4 1 4 1 4

- u v + - u w + - v w - -- u - -- v - -- w

8 8 8 16 16 16

> u2:=-1/2*s^2+1/3*r^2+8/3*t^2; v2:=-5/3*t^2+5/2*s^2+2/3*r^2;

> w2:=-(1/2)*r^2+3*t^2+(27/8)*s^2;

1 2 1 2 8 2

- - s + - r + - t

2 3 3

5 2 5 2 2 2

- - t + - s + - r

3 2 3

1 2 2 27 2

- - r + 3 t + -- s

2 8

> H2:=expand(-1/16*u2^2-1/16*v2^2-1/16*w2^2+1/8*u2*v2+1/8*u2*w2+1/8*v2*w2);

49 2 2 49 2 2 49 2 2 49 4 49 4 441 4

-- t r + -- t s + --- r s - -- t - --- r - ---- s

72 32 128 36 576 1024

> simplify(H1/H2);

576

---

49

>

Good idea leaving the square roots aside until the last step!

I did this on the two Heron's formula results only and got a better simplified expression for the ratio between the areas:

Sabc a b c

------ = -----------------------------------

Sdef a(b - r)(c - s) + t(c r + b(s - c))

Making r = b/2, s = c/3 and t = a/4 per the original problem, the expression easily evaluates to 24/7. Perhaps r should be associated with side a, s to side b, t to side c and the expression rewritten accordingly for mnemonic purposes.

This is exactly the approach I thought of since the beginning. I decided to use a triangle with sides proportional to 2, 3 and 4 as I realized the shape didn't matter (so did SlideRule as well). This simplified the solution, but required Heron's formula twice, the one for the inner triangle being a bit more laborious to solve by hand. Since the shape was irrelevant, why not to choose an even more convenient one? Indeed this allowed for an even more immediate solution. I still don't understand all the fuss about the use of these particular shapes, since they would have spared lots of tedious calculations as we've seen :-)