mersenneforum.org Count the Triangles
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

 2008-01-02, 20:01 #1 davar55     May 2004 New York City 423510 Posts Count the Triangles Start with a square. Divide each side into 5 equal parts, and draw the lines to form a 5x5 grid. Then cross-hatch the grid by drawing the diagonals of each of the 25 small squares. How many triangles can be traced in the resultant figure?
 2008-01-03, 02:46 #2 Jamiaz     Sep 2005 Detroit, MI 23 Posts I'm going to give it a guess here. Probably off, but it is worth a shot. 440 If it's correct or anyone wants to know I'll tell you how I came up with that number.
 2008-01-03, 03:31 #3 davieddy     "Lucan" Dec 2006 England 2·3·13·83 Posts I guess you are well on the low side
 2008-01-03, 04:16 #4 wblipp     "William" May 2003 New Haven 2,371 Posts It doesn't look that far off to me. I found 480. I initially got the same answer as Jaimaz, but when I checked my method on a 2x2 grid I found a few more triangles.
2008-01-03, 12:08   #5
davieddy

"Lucan"
Dec 2006
England

2×3×13×83 Posts

Quote:
 Originally Posted by wblipp It doesn't look that far off to me. I found 480. I initially got the same answer as Jaimaz, but when I checked my method on a 2x2 grid I found a few more triangles.
Yes I see what you mean. There are 55 squares with sides parallel to
the original each contributing 8 distinct triangles. But there are additional
triangles (I count 4 in your 2x2 grid) but can't generalize these rogues
ATM.

 2008-01-03, 16:06 #6 davieddy     "Lucan" Dec 2006 England 194A16 Posts I get 492
 2008-01-03, 16:34 #7 wblipp     "William" May 2003 New Haven 2,371 Posts Oh I see! There were still more with the hypotenuse parallel to a side. I now also find 492.
 2008-01-04, 11:49 #8 davieddy     "Lucan" Dec 2006 England 145128 Posts Apart from using symmetry to assert that there are an equal number of each of the 4 orientations of a given size of triangle, I don't see much potential for a smart answer to the question. There are two classes of triangles: hypotenuse parallel to sides or diagonals, with 5 sizes in each class. Admittedly the counting of each size within a class is similar but that still leaves us with 10 multiplications to perform. Anyone care to find a general solution?
2008-01-04, 12:23   #9
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by wblipp Oh I see! There were still more with the hypotenuse parallel to a side. I now also find 492.
I'm intrigued as to what you missed

 2008-01-05, 05:50 #10 wblipp     "William" May 2003 New Haven 2,371 Posts I think there will be separate solutions for odd and even numbers of squares. For the triangles with sides parallel to the square sides, in any one orientation, the "size 6-k" triangles can have base on one of k rows and located in one of k positions on that row, so there are 52+42+32+22+12 triangles with obvious generalization. For the triangles with hypotenuse parallel to the square sides, it's tricky because the height of a triangle is n/2, but the base can only be on integer. So the number of triangles of each size (in any one orientation) is Size 1: 5 rows x 5 positions. Size 2: 5 rows x 4 positions Size 3: 4 rows x 3 positions Size 4: 4 rows x 2 positions Size 5: 3 rows x 1 position This also has an easy generalization. The general solution needs to find the closed form sum for each of the generalizations and to multiply by 4 for the symmetric orientations. My error was in the number of rows. I had it decreasing too fast. William
2008-01-05, 10:25   #11
Richard Cameron

Mar 2005

2·5·17 Posts

Quote:
 Originally Posted by wblipp I think there will be separate solutions for odd and even numbers of squares. ...The general solution needs to find the closed form sum for each of the generalizations and to multiply by 4 for the symmetric orientations... and even: My error was in the number of rows. I had it decreasing too fast. William
I -presumably- went through the same reasoning as William, including annoyingly making a similar error (I got 4*121=484).
I know this isn't in the spirit of most puzzlers, but for Iggies and other lazy folks, a closed form is given at OEIS:

A100583: Number of triangles in an n X n grid of squares with diagonals

Richard

 Similar Threads Thread Thread Starter Forum Replies Last Post a nicol Math 1 2016-11-08 05:32 bcp19 Hardware 12 2013-04-26 13:52 Moose Lounge 5 2012-03-09 03:22 davieddy Puzzles 15 2007-04-06 20:16 Citrix Puzzles 15 2005-09-11 18:03

All times are UTC. The time now is 13:23.

Fri May 27 13:23:35 UTC 2022 up 43 days, 11:24, 1 user, load averages: 1.32, 1.65, 1.79

Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔