Lost In A Square
An ant is lost in a square, and his distances to the vertices of the square are 7, 35, 49, and x . Find x .
Note: The image is not drawn to scale.
The answer is 35.
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5 solutions
(This comment is less relevant, because I missed the condition that the Dungeon is a square.)
There is a much more direct method via Pythagorean theorem, which tells us that 7 2 + 4 9 2 = 3 5 2 + x 2 .
This is also known as the British Flag Theorem.
By right, you will also need to state the order that the distances are in, since we could have P A = 7 , P B = 4 9 , P D = 3 5 , which will yield P C 2 = 3 5 7 7 . I've added in the explicit restriction that x is an integer.
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But if PC squared = 3577 would this yield a square?
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Oh yes, I missed that condition. Having it be a square will certainly place further restrictions on it. My point was that you assumed that P A = 7 , P B = 4 9 , P D = 3 5 , which may not necessarily be the case. You will still need to show that the other permutations do not lead to a square shaped dungeon. I've edited the question back to the original form
Yup, corrected the cube. Thanks!
I did it by the same method sir but I didn't knew about this theorem, just found it accidentally when I was approaching this question using coordinate geometry, it made it simpler just by solving three simple equations I got this formula ......thank you for ur illumination on this...!
35 cubed should be 35 squared, shouldn't it?
Thanks for the Theorem, Calvin.
Oh sir calvin I only thought that BFT is only used in a rectangle.. I didnt know it can also be applied in the square.. Thanks :D
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Squares are a subset of rectangles, so BFT applies.
The reason why I said it is less relevant, is because I was first raising up the issue that there could be possible answers for x , but I didn't realize that we had a square (See Guiseppi's reply).
Nice problem sir Butel
Well, there is a relation (I really don't remember the name, I once had to prove it in a math class) that states the following
Square A B C D ( A opposing C and B opposing D ) and point P anywhere inside A B C D
It is true that ( A P ) 2 + ( D P ) 2 = ( C P ) 2 + ( B P ) 2
Now, using that info, the problem is really easy
therefore we have three possibilities
1- 3 5 2 + x 2 = 4 9 2 + 7 2
2- 4 9 2 + x 2 = 3 5 2 + 7 2
3- 7 2 + x 2 = 3 5 2 + 4 9 2 .
And, to simplify the solution we seek the integer solution to this, and som the answer is 35
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It's called the british flag theorem .
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Thank you! I have been trying to find it all day long!
The sol that has x=28sqrt2 can't work if you consider triangle APD AP=35 DP=1 so then length x must be between 35-1< x < 35+1 to make it a closed triangle. 28sqrt2=39.6 so can not work. 21sqrt2=29.7 also can't work.
What am i missing?
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D P = 7 and not 1. So we have ( 28 < x < 42 ( instead.
7²+49²= 35²+ x² ~ x = 35
Dependent upon the distribution of the vertices within the square three different equations can be derived:
x(2) + 7(2) = 35(2) + 49(2) Where x = 59.80802622
x(2) + 35(2) = 7(2) + 49(2) Where x = 35
x(2) + 49(2) = 7(2) + 35(2) Where x = 33.57082066i (This is an imaginary number)
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What's the length of the square for x = 5 9 . 8 ?
(I made the same mistake by wanting to apply the British Flag Theorem, but that applies to rectangles, and so we have to verify if we are actually in a square.)
Considerably more complicated than the solution given above but I never heard of the British flag theorum until now. Still, I enjoyed solving this particular problem, after scratching my head for a while 😂😂😂😂😂😂
Applying British Flag Theorem
x = sqrt( 7^2 + 49^2 -35^2) x= sqrt( 1225) x= 35
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I wish someone had told me about the British Flag Theorem!! It would have made life so much simpler!!!!!
I appreciate the neatness of the solution :)
By the British Flag Theorem, 7² + 49² = x² + 35² which readily yields, x = 35.
But how do you know that the above is the correct combination?
No complicated calculation, nor no need to know any theorem. Let me use the diagram drawn by Guiseppi Butel. Let a = length of the side of the square, b = (horizontal length of DP along DC), c = (vertical length of DP along DA). Then
[1] b 2 + c 2 = 7 2
[2] b 2 + ( a − c ) 2 = 3 5 2
[3] ( a − b ) 2 + ( a − c ) 2 = 4 9 2
[4] ( a − b ) 2 + c 2 = x 2
In the above set of equations, [1] + [3] = [2] + [4] (this is the British Flag theorem), so x = ( 4 9 2 + 3 5 2 − 7 2 ) = 3 5 .
Let P be the point of intersections in the square. Let u and w be lines drawn from P to the sides of the square(perpendicular to each other. Then if the side of the square is S, we have the following equations using the Pythagorean Theorem:(1) u^2 + (S - w)^2 = 49. (2) (S - u)^2 + (S - w)^2 = 1225. (3) (S - u)^2 + w^2 = 2401. (4) u^2 + w^2 = x^2. Subtracting (1) from (2), (S - u(^2 - u^2 = 1176. Subtracting (2) from (3), w^2 - (S - w)^2 = 1176. Therefore, (S - u(^2 - u^2 = w^2 - (S - w)^2. Factoring the difference of 2 squares, (S - 2u)(S) = (2w - S)(S). Dividing by S, S - u = w. Substituting into (3), 2w^2 =2401, and w^2 = 1200.5. Substituting into (1), 2u^2 = 49, and u^2 = 24.5. Then u^2 + w^2 = x^2 = 1225, and x = 35. Ed Gray
Place S at point P.