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 March 22nd, 2014, 07:26 PM #1 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 which one is the lengthiest $\triangle ABC ,\,\,$ with side length : $a,b,\,\, and \,\, c$ we have : $(1) a^2-a-2b-2c=0$ $(2)a+2b-2c+3=0$ now determine among : $a,b,c$ which one is the lengthiest
 March 22nd, 2014, 08:13 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Re: which one is the lengthiest You want to express b in terms of a only and c in terms of a only. You can do this by eliminating b and c in turn from the equations in the same manner that you would to solve a system of simultaneous equations. You should then be able to order them appropriately. I made $c > b$ for all values of $a$ and $a > c$ when $1 \lt a \lt 3$.
 March 22nd, 2014, 08:23 PM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: which one is the lengthiest c. Also, if a,b,c are integers, then b>a except for one case: (a,b,c) = (5,3,7)
 March 22nd, 2014, 09:26 PM #4 Senior Member   Joined: Mar 2014 Posts: 112 Thanks: 8 0 = a² - 4c + 3 i.e. 4c = a² + 3. Hence c = (a² + 3)/4. 0 = a² - 2a - 4b - 3 i.e. 4b = a² - 2a - 3 = (a - 3)(a + 1). Hence b = (a - 3)(a + 1)/4. Since b > 0 is correct, a > 3 must also be correct. If we use the solution Denis gave we are done.
March 24th, 2014, 09:14 PM   #5
Math Team

Joined: Oct 2011

Posts: 14,597
Thanks: 1039

Re: which one is the lengthiest

Quote:
 Originally Posted by Albert.Teng $(2)a+2b-2c+3=0$
Was diddling with this "Albert Special" during the hockey game...

Above equation can be rewritten as:
2c - 2b = a + 3 [1]
So evident that c > b (since a,b,c are all > 0, being triangle sides)

Can't find "as simple" a way to show c > a...

 March 25th, 2014, 11:24 AM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: which one is the lengthiest Albert, something for your "useless information(!)" file; here's the first 8 integer solutions: Code: n a b c 1 5 3 7 2 7 8 13 3 9 15 21 4 11 24 31 5 13 35 43 6 15 48 57 7 17 63 73 8 19 80 91 a = 2(n - 1) + 5 b = n(n + 2) c = n^2 + 3n + 3 You're welcome; no charge :P

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