MHB Prove Every Room w/ Odd Doors in House Has a TV Set - H for American Buyer

  • Thread starter Thread starter Chris11
  • Start date Start date
  • Tags Tags
    Fun Puzzle
AI Thread Summary
In a house with only one entrance and a maximum of one door between any two rooms, the owners place a TV in every room with an odd number of doors to appeal to an American buyer. Using the Handshaking lemma, it is established that if all rooms had an even number of doors, it would contradict the existence of a single entrance with an odd degree. Therefore, at least one room must have an odd number of doors, ensuring that it contains a TV. As the buyer evaluates the house, they will inevitably find a TV in at least one room. This logical deduction confirms that the buyer will successfully locate a TV set during their visit.
Chris11
Messages
26
Reaction score
0
Let H be a house with only one entrence. Assume that between any two rooms in the house there can only be at most one door. The owner's of this house, keen on selling to an American buyer, decide to place a TV set in every room with an odd number of doors.

Prove that the the American will be able to find a TV set when he comes to evaluate the house.
 
Mathematics news on Phys.org
Chris11 said:
Let H be a house with only one entrence. Assume that between any two rooms in the house there can only be at most one door. The owner's of this house, keen on selling to an American buyer, decide to place a TV set in every room with an odd number of doors.

Prove that the the American will be able to find a TV set when he comes to evaluate the house.

Hi Chris11, :)

I think we can use the Handshaking lemma to solve this problem. Suppose every room in the house has even number of doors. Represent rooms by vertices and doors by edges and we get a finite graph (since the number of rooms should be finite). We shall consider the garden (outside) also as a vertex. The "garden vertex" has degree 1 (only one entrance), and all the other vertices have even degrees. According to the Handshaking lemma this cannot happen. Therefore there should be at least one vertex (a room) except the "garden vertex" with odd degree (odd number of doors). This room contains a TV and if the buyer walks into each room he should find this.

Kind Regards,
Sudharaka.
 
Yep! That was my solution.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Back
Top