Author |
Topic: Two Triangles (Read 518 times) |
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender: 
Posts: 4489
|
 |
Two Triangles
« on: Apr 6th, 2004, 9:33am » |
 Quote  Modify
|
Let ABC be a primitive right triangle with integer sides and integer area.
Let DEF be an isosceles triangle with integer sides and integer area.
If ABC and DEF have the same perimeter and the same area, find the lengths of the sides of ABC and of DEF.
|
|
 IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
Sir Col
Uberpuzzler
impudens simia et macrologus profundus fabulae
Gender:
Posts: 1825
|
|
Re: Two Triangles
« Reply #1 on: Apr 6th, 2004, 4:59pm » |
Quote Modify
|
I don't have a solution, but I've made a start (albeit fairly obvious)...
::
Let the sides of a right angle triangle are given by x=2mn, y=m2–n2, and z=m2+n2 (hypotenuse).
P=2m(m+n), and A=mn(m+n)(m–n); so A=P(m–n)/2.
As the right angle triangle is a primitive, one of m and n must be odd and the other even, otherwise all the sides would be even.
For the isosceles triangle to have integer sides and integer area, the base*height/2 must be integer, which means that it could be two Pythagorean right angle triangles back-to-back; although this is not necessary: the base of each right angle triangle could be a multiple of 0.5 and the height a multiple of 2.
Assuming that the three sides of the right angle triangle are given by, a=2jk, b=j2–k2, and c=j2+k2 (hypotenuse).
Depending on orientation of the right angle triangles, the sides of the isosceles will be c-b-c or c-a-c.
P1=2(j2+k2)+2(j2–k2)=4j 2
P2=2(j2+k2)+2*2jk=2(j+k)2
But whichever orientation, A=2jk(j+k)(j–k).
From here, I'm unable to establish anything else.
::
|
| « Last Edit: Apr 6th, 2004, 5:11pm by Sir Col » |
IP Logged |
mathschallenge.net / projecteuler.net
|
|
|
SWF
Uberpuzzler
Posts: 879
|
|
Re: Two Triangles
« Reply #2 on: Apr 7th, 2004, 5:06pm » |
Quote Modify
|
Answer:
Using expressions similar to Sir Col's for Pythagorean triples, also find the area in terms of perimeter with Heron's formula. For the isosceles triangle with base q, and leg r:
16*A*A=p*(p-2r)*(p-2r)*(p-2q)
This implies p*(p-2q) is a perfect square,...
Right triangle: a=135, b=352, c=377
Isosceles: q=132 r=366
Perimeter=864 Area=23760
.
|
|
IP Logged |
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Two Triangles
« Reply #4 on: Apr 8th, 2004, 5:51am » |
Quote Modify
|
A very nice problem  ! I just wanted to post my solution, but saw SWF and NickMcG came first 
What about the following variation:
Find a pair of primitive isosceles Heronian triangles with the same perimeter and area.
Does a triple of such triangles exist?
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print