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| All rationals? (Posted on 2005-12-13) |
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A square ABCD is incribed in a circle. Prove that the distances between a point P on the circumference to the four vertices of the square can´t all be rational numbers.
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Submitted by pcbouhid
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Rating: 3.0000 (2 votes)
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Solution:
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Let L be the side of the inscribed square and WLOG suppose that P is in the arc AB.
If P coincides with B, then PA = PC = L and PD = L√2 . Thus, if L is rational then PD is irrational, and if L is irrational then PA and PC are irrationals.
Analogously, if P coincides with A.
Suppose now that P is between A and B and that PA, PB, PC and PD are all rationals. The quadrilateral PBCD is inscribled in the circle. Thus, applying Ptolomeu´s theorem, we have:
PD * BC + PB * CD = BD * PC. We have BC = CD = L and BD = L√2. Therefore PD + PB = PC√2.
But PD and PB rationals imply PD + PB rational, and PC rational imply PC√2 irrational.
That is, a rational number is equal to an irrational number. What is a contradiction that occurred with the initial hypothesis that PA, PB, PC, PD were all rationals.
The conclusion is that at least one of them have to be irrational. |
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