Author |
Topic: Unique Isosceles Triangle (Read 1398 times) |
|
Sir Col
Uberpuzzler
impudens simia et macrologus profundus fabulae
Gender: 
Posts: 1825
|
 |
Unique Isosceles Triangle
« on: Apr 13th, 2005, 4:33pm » |
 Quote  Modify
|
(i) An isosceles triangle has a perimeter of 12 and an area of 4sqrt(3). Given that the lengths of the sides are integer in this case, find the unique solution for the dimensions of the sides.
(ii) If the perimeter of an isosceles triangle is P and the area is A, prove that there can be no more than two solutions.
(iii) Find A, in terms of P, for which there is a unique solution.
|
| « Last Edit: Apr 14th, 2005, 12:37am by Sir Col » |
 IP Logged |
mathschallenge.net / projecteuler.net
|
|
|
markr
Junior Member
Gender:
Posts: 90
|
|
Re: Unique Isosceles Triangle
« Reply #1 on: Apr 13th, 2005, 9:57pm » |
Quote Modify
|
(1)I get a third degree equation with roots 4, 4, 7. 7 is too large, so the answer is 4.
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: Unique Isosceles Triangle
« Reply #2 on: Apr 14th, 2005, 2:29am » |
Quote Modify
|
| hidden: |
Let's call the two equal sides a, and the other one b. So the perimiter p=2a+b
Then we have
A = b/4 sqrt(4a^2 - b^2))
A = b/4 sqrt((2a+b)(2a-b))
A = b/4 sqrt(p (p-2b))
4 A = b sqrt(p (p-2b))
16 A^2 = b^2p^2-2b^3p
So we get the cubic
b^3 - p b^2/2 + 8 A^2/p = 0
Assuming it is of the form *)
(b-x)^2(b-y)=0
b^3 - b^2 (2 x + y) + b(x^2 + 2xy ) - x^2y=0
We can solve it from the following three equations
2x + y = p/2
x^2 + 2xy = 0 => x + 2y = 0
8 A^2/p = - x^2 y
p = 3x -> x = p/3 {b=x, means all sides are 1/3 of the perimeter, so it's an equilateral triangle}
y = -1/6 p {b=y means we have a triangle with a negative side to it, so that's nto really an option}
8 A^2/p = - (p/3)^2 (-1/6 p)
So A = p^2 sqrt(3)/36
|
*) probably not a fair assumption, but it sure makes the calculation easier 
|
| « Last Edit: Apr 14th, 2005, 2:32am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Unique Isosceles Triangle
« Reply #3 on: Apr 14th, 2005, 5:25am » |
Quote Modify
|
| hidden: | Let P be fixed, and consider A as a function of the altitude onto the base h. Clearly, A(h=0) = A(h=P/2) = 0.
What can we say about this function's behaviour when h goes from 0 to P/2 ?
|
|
|
IP Logged |
|
|
|
Deedlit
Senior Riddler
Posts: 476
|
|
Re: Unique Isosceles Triangle
« Reply #4 on: Apr 15th, 2005, 4:49pm » |
Quote Modify
|
You can also solve it with Lagrange multipliers.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print