(1) Just for a warmup, a very old logic puzzle: (If you have seen this before, don’t spoil it by telling the answer too soon.) (tricky answer)
An Arab rides into an oasis and sees a group of people sitting and arguing. He parks his camel at the water trough and asks what the problem is. It seems that the father of the family has died and they are having trouble putting his will through probate. The will says “To my oldest son, who will manage the family caravan business, I leave half my camels. To my second son, who has a family to support, I leave a third of my camels. To my youngest son, I do not want to forget you, so I leave you one-ninth of my camels.” The visitor asks what the big hold-up is. “But sir, he left seventeen camels!” The sons are distraught; they fear that out of respect for their father they will have to sell everything to the used camel lot and split the money. The visitor thinks for a while, and then gives them a solution that allows evceryone to ride happily into the sunset. What is the solution?
(2) A second old logic puzzle, easier:
A man goes to market and comes back with a fox, a goose, and a bag of grain. He has to cross the river in a small rowboat, and he can only take one thing at a time — the grain is heavy, and he can’t both row and keep the fox and goose apart. So he is going to take one thing and leave two on shore for the next trip. But if he leaves the goose with the grain, it will break open the bag and eat it. If he leaves the fox with the goose, the goose will be supper. How does he get across, and what is the minimum number of trips?
(3) A geometry puzzle (fairly difficult):
It is easy to prove that, if a triangle is isosceles, the medians of the equal sides are equal. Now, for the hard part, prove the converse: if two medians of a triangle are equal, prove that the triangle is isosceles.
(Be very careful not to assume anything else except that the medians are equal)
Hint — look for a theorem about medians in an old geometry text.
(4) Another geometry puzzle (quite difficult)
Draw any triangle ABC. Extend the base BC in both directions to R and S to form two exterior angles, RBA and SCA. Bisect the exterior angles RBA and SCA. Also bisect the original interior angles ABC and ACB. Extend these bisectors until they meet at P and Q. The bisector of RBA will be BP, meeting the bisector of ACB, CP; and the bisector of SCA will be CQ, meeting the bisector of ABC, BQ. Finally join the straight lines PA and QA.
Prove that PAQ is a straight line.
This is difficult, although it can be done with the theorems in the old high school text. You may find a useful direction (not a solution, but guidance towards one) in a theorem about bisectors in the old geometry text.
(warning - avoid unjustified assumptions and circular arguments.)
Re: To Sara McName (and others) -- more math puzzlers
Thanks for not posting the two oldies — I’m hoping to see some creative ideas.
The angle bisector problem can probably be done with trig identities — I haven’t done it that way — but in fact it solves quite reasonably using only pure Euclid, material in the old high school text, although some high-level logic needed. Try it and see!
More math puzzlers that a child can solve . .
These require no high level math skills, just (very) creative thinking:
(5) Given six toothpicks, without breaking or bending or otherwise changing them in any way, form (simultaneously) four congruent triangles.
(6) Without lifting your pen, draw four connected straight line segments that cross each of these points:
o o o
o o o
o o o
(7) Without lifting your pen, draw three connected straight line segments that cross each of these:
O O O
O O O
O O O
I’m trying to find the “hotel problem” but it’s so complex I have to look it up — more later.
Re: To Sara McName (and others) -- more math puzzlers
This is a geometric solution to problem 4 in the original post.
Angles PBQ and PCQ are right angles, so triangles PBQ and PCQ are right triangles that share PQ as a diameter. That means that PBCQ is a circle. Angles BPC and BQC both measure the arc BC so BPC equals BQC. The measure of the apex angle ABC is twice the measure of the inscribed angle BPC, so angle ABC is a central angle of circle PBCQ, and point A is the center of circle PBCQ. PAQ is the diameter of PBCQ and is therefore a straight line.
I’ve tried to do this with trig indenties. I was good practice, but I didn’t find a trigonometric solution.
Sara McName
hmmm. . . .
This is close to a proof. You don’t justify why the apex angle, BAC (typo) is twice BPC or BQC, but I worked it out and yes, OK.
Trouble is, that makes A the center — IF A is on the line PQ. But you can make the same angle without it being on that line, and the fact that A is on the straight line PQ is the hypothesis that was to be proved. Darn!
When I originally saw this, a proof was resented that used all sorts of circles, but it was horrendously complicated and messy. I worked on it for a while and developed a proof that is much simpler; uses a couple of uniqueness arguments.
There probably is sa method to do with trig and/or vectors, but also probably very messy. I’m going to think about it.
oh-oh! Amazing theorem proved!! Trouble is, it's false. . .
Looking again at my notes from your work in the light of day after a nagging feeling last night, I see this:
We start with *any* triangle ABC.
We go through the construction of the bisectors etc.
You prove that B, C, P, Q are on a circle.
I prove that A is indeed on PQ (I can give you my proof later)
You prove and I check that the apex angle BAC is indeed twice angles BPC and BQC, so BAC should be the central angle, and voila, A is the center of the circle.
New conclusions to be reached from this: AB and AC are both radii of the circle. So AB = AC in *any* triangle.
**All triangles are isosceles!**
Repeating the act with the third vertex, BA = BC in *any* triangle
**All triangles are equilateral!**
We have reached a very interesting but possibly slightly overgeneralized conclusion. This calls for the famous request for more money and Further Research.
****************************************
Amusing anecdote that this reminds me of: in a math education class, we did a project, and some of us chose to videotape a convenient child. I did daughter and friend, age 7-8, puzzling over area problems. You could *see* the wheels turning in the heads.
Anyway, another student taped a kid age 4 who had clearly been very well trained in a well-intentioned but possibly limited daycare. She had him identify shapes. She showed him square and circle and he did fine. She showed him an equilateral triangle and he confidently identified it as a triangle. Then she showed him a right-angled triangle, standard positioned with legs vertical and horizontal, and he puzzled over it for quite a while and said “Half a triangle?”
found it -- tell me when you've finished puzzling with this
I hacked at this a while longer and found what is missing. Fatal flaw in your proof, I fear. When you’ve finished let me know and I’ll post/send what I’ve got.
Re: found it -- tell me when you've finished puzzling with t
I know my proof is not really a proof. I thought it might draw others into the discussion.
Re: To Sara McName (and others) -- more math puzzlers
Sorry to be so long in getting back to this, but I’m old and sometimes slow. I showed the straight line problem to the math teachers at my school. One of them assigned it as a construction for extra credit, but none have attempted a solution. And no one here on the board seems to be interested in geometry either.
I think one possible proof may go as follows. PBCQ is a circle with PQ as the diameter because angle PBQ and angle PCQ are right angles. Next one must extend lines CA and BA until they intersect the circle as D and E.
Then one could show that PQ is the bisector of angle DPB because equal arcs have equal inscribed angles. Then because DEA is congruent to BCA, the intersection of BA and DA is also a bisector of angle DPB.
Probably one could show that an angle can have only one bisector, so PA and PQ must be the same straight line.
Re: To Sara McName (and others) -- more math puzzlers
Skip the circles, which lead into many interesting but dead-end byways.
Your last sentence, the uniqueness of bisectors, is the key to my neat proof. I am too tired now to check in detail but think you need something more to prove that you do indeed have bisectors.
I saw this problem in a book and the proof there was awful, all sorts of exscribed circles and arcane theorems.
The bisectors work well if you remember one nice theorem about bisectors and triangles and then extend it.
Re: To Sara McName (and others) -- more math puzzlers
Probably I should not post an answer to the first two problems because I’ve seen them before. If I remember correctly, it was the day my brother was mad at me for wearing a Landon button. That was Alf, not Michael. And it was mules not camels.
The fourth problem about the bisecting of the angels may be a good geometry problem, but it is an even better trig problem. Lots of practice with trig identities.
Re: To Sara McName (and others) -- more math puzzlers
Thanks for not posting the two oldies — I’m hoping to see some creative ideas.
The angle bisector problem can probably be done with trig identities — I haven’t done it that way — but in fact it solves quite reasonably using only pure Euclid, material in the old high school text, although some high-level logic needed. Try it and see!
More math puzzlers that a child can solve . .
These require no high level math skills, just (very) creative thinking:
(5) Given six toothpicks, without breaking or bending or otherwise changing them in any way, form (simultaneously) four congruent triangles.
(6) Without lifting your pen, draw four connected straight line segments that cross each of these points:
o o o
o o o
o o o
(7) Without lifting your pen, draw three connected straight line segments that cross each of these:
O O O
O O O
O O O
I’m trying to find the “hotel problem” but it’s so complex I have to look it up — more later.
Re: To Sara McName (and others) -- more math puzzlers
This is a geometric solution to problem 4 in the original post.
Angles PBQ and PCQ are right angles, so triangles PBQ and PCQ are right triangles that share PQ as a diameter. That means that PBCQ is a circle. Angles BPC and BQC both measure the arc BC so BPC equals BQC. The measure of the apex angle ABC is twice the measure of the inscribed angle BPC, so angle ABC is a central angle of circle PBCQ, and point A is the center of circle PBCQ. PAQ is the diameter of PBCQ and is therefore a straight line.
I’ve tried to do this with trig indenties. I was good practice, but I didn’t find a trigonometric solution.
Sara McName
hmmm. . . .
This is close to a proof. You don’t justify why the apex angle, BAC (typo) is twice BPC or BQC, but I worked it out and yes, OK.
Trouble is, that makes A the center — IF A is on the line PQ. But you can make the same angle without it being on that line, and the fact that A is on the straight line PQ is the hypothesis that was to be proved. Darn!
When I originally saw this, a proof was resented that used all sorts of circles, but it was horrendously complicated and messy. I worked on it for a while and developed a proof that is much simpler; uses a couple of uniqueness arguments.
There probably is sa method to do with trig and/or vectors, but also probably very messy. I’m going to think about it.
oh-oh! Amazing theorem proved!! Trouble is, it's false. . .
Looking again at my notes from your work in the light of day after a nagging feeling last night, I see this:
We start with *any* triangle ABC.
We go through the construction of the bisectors etc.
You prove that B, C, P, Q are on a circle.
I prove that A is indeed on PQ (I can give you my proof later)
You prove and I check that the apex angle BAC is indeed twice angles BPC and BQC, so BAC should be the central angle, and voila, A is the center of the circle.
New conclusions to be reached from this: AB and AC are both radii of the circle. So AB = AC in *any* triangle.
**All triangles are isosceles!**
Repeating the act with the third vertex, BA = BC in *any* triangle
**All triangles are equilateral!**
We have reached a very interesting but possibly slightly overgeneralized conclusion. This calls for the famous request for more money and Further Research.
****************************************
Amusing anecdote that this reminds me of: in a math education class, we did a project, and some of us chose to videotape a convenient child. I did daughter and friend, age 7-8, puzzling over area problems. You could *see* the wheels turning in the heads.
Anyway, another student taped a kid age 4 who had clearly been very well trained in a well-intentioned but possibly limited daycare. She had him identify shapes. She showed him square and circle and he did fine. She showed him an equilateral triangle and he confidently identified it as a triangle. Then she showed him a right-angled triangle, standard positioned with legs vertical and horizontal, and he puzzled over it for quite a while and said “Half a triangle?”
found it -- tell me when you've finished puzzling with this
I hacked at this a while longer and found what is missing. Fatal flaw in your proof, I fear. When you’ve finished let me know and I’ll post/send what I’ve got.
Re: found it -- tell me when you've finished puzzling with t
I know my proof is not really a proof. I thought it might draw others into the discussion.
Re: To Sara McName (and others) -- more math puzzlers
Sorry to be so long in getting back to this, but I’m old and sometimes slow. I showed the straight line problem to the math teachers at my school. One of them assigned it as a construction for extra credit, but none have attempted a solution. And no one here on the board seems to be interested in geometry either.
I think one possible proof may go as follows. PBCQ is a circle with PQ as the diameter because angle PBQ and angle PCQ are right angles. Next one must extend lines CA and BA until they intersect the circle as D and E.
Then one could show that PQ is the bisector of angle DPB because equal arcs have equal inscribed angles. Then because DEA is congruent to BCA, the intersection of BA and DA is also a bisector of angle DPB.
Probably one could show that an angle can have only one bisector, so PA and PQ must be the same straight line.
Re: To Sara McName (and others) -- more math puzzlers
Skip the circles, which lead into many interesting but dead-end byways.
Your last sentence, the uniqueness of bisectors, is the key to my neat proof. I am too tired now to check in detail but think you need something more to prove that you do indeed have bisectors.
I saw this problem in a book and the proof there was awful, all sorts of exscribed circles and arcane theorems.
The bisectors work well if you remember one nice theorem about bisectors and triangles and then extend it.
Probably I should not post an answer to the first two problems because I’ve seen them before. If I remember correctly, it was the day my brother was mad at me for wearing a Landon button. That was Alf, not Michael. And it was mules not camels.
The fourth problem about the bisecting of the angels may be a good geometry problem, but it is an even better trig problem. Lots of practice with trig identities.