All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Eight distances (Posted on 2014-12-10) Difficulty: 2 of 5
Within the top left-hand quarter of a square(its side an integer below 40 cm) I marked two points, each of which was distant an integral number of cm, not only from each side of the square but also from both the top left-hand corner and the bottom right-hand corner of the square.

What size was a side of this square?

Based on Enigma problem (New Scientist 2010)

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution answer | Comment 2 of 4 |
Given no restriction on side length other than it is less than 40 cm, there are at three possible sizes: 24, 33, and 36.
With the top-left corner being designated as (0,0) and the bottom-right corner of the square as (x,x), such that x is equal to the squares side length, the marked points in the top-left corner will be, respectively:
(3,4) & (4,3), (5,12) & (12,5), and (8,15) & (15,8).

Given the amended restriction that the side is of odd length, the solution is then 33 cm, with the two marked points being (5,12) and (12,5).

As a point marked must be an integral length from each corner and each side of the square, the marked point will be a vertex to two pythagorean triangles,  and an end point of each of those triangle's hypotenuse. The short leg of one when added to the long leg of the other will equal in length to the total length of other two legs, and to the length of the side of the square. 

As the marked points are in the top-left quadrant of the square, the hypotenuse of the pythagorean triangle will be less than or equal to 26 cm, and no leg will be longer than 19 cm. This leaves only the four primitive pythagorean triples (3,4,5), (5,12,13), (8,15,17), (7,24,25) and two non-primitive pythogarean triples (6,8,10) and (9,12,15) to consider. 
The complementary pythagorean triple must then have a hypotenuse less than the floor(sqrt((39-L1)^2+(39-L2)^2)), where L1 and L2 are the legs of the other triple. This gives the triples (3,4,5), (6,8,10), (9,12,15), (12,16,20), (15,20,25), (18,24,30), (21,28,35), (5,12,13), (10,24,26), (15,36,39), (8,15,17), (16,30,34), (7,24,25), (20,21,29), (9,40,41), (12,35,37) to consider.
The long leg of the top-left and the short leg of the bottom-right must equal that of the short and long leg of the others, and with the addendum, the total must be odd. In addition, the two triples can not be the same (else it would violate the condition that the marked points were in the top-left quadrant). This leaves the two triples to be (5,12,13) and (21,28,35) as the only possibility.
5+28 = 12+21 = 33. The side-length of the square is 33 cm.

Edited on December 10, 2014, 3:43 pm
  Posted by Dej Mar on 2014-12-10 11:17:10

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information