A circle is inscribed in a hexagon, as shown in the diagram.
Is it possible that the side lengths of the hexagon are 7 , 9 , 1 1 , 1 3 , 1 5 , 1 7 in some order?
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This problem is extremely similar in concept to the first problem from the Basic set for this week . In that problem, we wanted to know if it was possible for x y z + y + z + x
to all be odd. With this problem, we want
x y z v w u + y + z + v + w + u + x
to all be odd. The extra condition that x + y + z + v + w + z is an odd number causes this case to fail.
Did you mean AB + CD + EF = x + y +z+ v + w + u (rather than z) = 36 ?
As always a very good solution. I solved it by taking its converse true and then reaching an obvious result..
How do you know that your beginning diagram is true and that two of the line segments don't overlap?
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I'm not sure what you mean here; we're supposing it's true. We also know the congruences x = x, y = y, etc. are true because both line segments are tangent to a circle from a point outside of it.
Same method, is this problem an original? It's nice
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I believed it's a revised version I saw in a recent math Olympiad test.
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Ah thanks, it's nice when problems combine geometry with number theory:)
where does the equation AB+CD+EF = half the circumference come from?
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Suppose yes. Then we will use the figure below.
x + y + z + v + w + u = 2 ( x + y ) + ( y + z ) + ( z + v ) + ( v + w ) + ( w + u ) + ( u + x ) = 2 7 + 9 + 1 1 + 1 3 + 1 5 + 1 7 = 3 6
From that A B + C D + E F = x + y + z + v + w + z = 3 6 which is impossible, since the sum of three odd integers is always an odd number. (Note: The 7 , 9 , 1 1 , 1 3 , 1 5 , 1 7 are all odd integers.)
Therefore it is impossible.