Teaser 2687: Creased
From The Sunday Times, 23rd March 2014 [link] [link]
I took a rectangular piece of paper whose sides were whole numbers of centimetres — indeed the longer side was exactly three times the length of the shorter side. I folded the rectangle along a diagonal. I then took the folded piece and folded it again along its line of symmetry. Then I completely unfolded it to see that the two creases had divided the rectangle into two equal triangles and two equal quadrilaterals. In square centimetres the area of each triangle was a three-figure number, and the area of each quadrilateral was another three- figure number using the same three digits but in a different order.
What was the area of each triangle?
[teaser2687]
GeoffR, and 
Hugh Casement are discussing. 
Frits are discussing. 
Brian Gladman are discussing. 
John Crabtree, and 
Mark Valentine are discussing. 
S2T2 
Jim Randell 9:14 am on 10 January 2023 Permalink |
If the dimensions of the original rectangle are x by 3x. Then it has an area of 3x².
The result of the folds looks like this:
X is in the centre of the rectangle (so 3x/2 from the left/right edges and x/2 from the top/bottom edges).
And Y is vertically below X, so we have:
The triangles OXY and XZY are similar. With the sides of OXY being 3× those of XZY.
Hence:
And the area of triangle OXZ is given by:
And the area of the quadrilateral ABXZ is:
The following Python program runs in 51ms. (Internal runtime is 99µs).
Run: [ @replit ]
from enigma import (irange, inf, nsplit, printf) # digits of a number, in order digits = lambda n: sorted(nsplit(n)) # consider x values for x in irange(1, inf): x2 = x * x (T, t) = divmod(x2 * 5, 12) (Q, q) = divmod(x2 * 13, 12) # T and Q are 3 digit numbers, with no remainder if Q > 999: break if t > 0 or q > 0 or T < 100: continue # do T and Q use the same digits? if digits(T) == digits(Q): printf("x={x}: T={T} Q={Q}")Solution: The area of each triangle is: 135 cm².
And the area of the quadrilaterals is: 351 cm².
The original rectangle had dimensions: 18 cm × 54 cm. (Area = 972 = 2 × (135 + 351)).
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GeoffR 1:22 pm on 10 January 2023 Permalink |
% A Solution in MiniZinc include "globals.mzn"; % using Jim's analysis and notations var 1..50:x; var 1..50:OX; var 1..50:OY; var 1..50:XZ; var 100..250:T; var 100..500:Q; constraint 2 * OX == 3 * x /\ 2 * OY == x; constraint 12 * T == 5 * x * x; constraint 12 * Q == 13 * x * x; % Digits in triangular area (a, b, c) var 1..9:a; var 0..9:b; var 1..9:c; constraint a == T div 100 /\ b == T div 10 mod 10 /\ c == T mod 10; % Digits in quadrilateral area (d, e, f) var 1..9:d; var 0..9:e; var 0..9:f; constraint d == Q div 100 /\ e == Q div 10 mod 10 /\ f == Q mod 10; % Digits in triangular area = Digits in quadrilateral area constraint {a, b, c} == {d, e, f}; solve satisfy; output ["Triangle area = " ++ show(T) ++ " sq cm" ++ "\n" ++ "Quadrilateral area = " ++ show(Q) ++ " sq cm"]; % Triangle area = 135 sq cm % Quadrilateral area = 351 sq cm % ---------- % ==========LikeLike
Jim Randell 9:40 am on 11 January 2023 Permalink |
@Geoff: I think the digit constraint at line 21 needs to be a bit stronger. Two numbers that had repeated digits could pass the test without being reorderings of the same collection of digits (e.g. 141 and 414).
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GeoffR 10:03 am on 11 January 2023 Permalink |
@Jim:
Maybe like this?
% Digits in triangular area = Digits in quadrilateral area constraint {a, b, c} == {d, e, f} /\ a != d /\ b != e /\ c != f /\ all_different([a, b, c]);LikeLike
Jim Randell 11:22 am on 11 January 2023 Permalink |
I think something like this would be more appropriate:
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Hugh Casement 7:46 am on 11 January 2023 Permalink |
As triangles OXY and XZY are similar, YZ = x/6, which gives us another way to calculate T
as half base OZ times height XY, without any square roots.
I notice that whenever two integers in the ratio 5:13 have the same digit sum,
that is 9 or a multiple of 9; e.g. 45:117, 90:234, 270:702.
The last of those would be a further solution, wouldn’t it?
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Jim Randell 8:58 am on 11 January 2023 Permalink |
@Hugh: T = 270, Q = 702 give x = 18√2.
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