bonanova

Everything in your post is correct. But remember what the question is. Whatever value a has, how close does the ray come to the point O? The phrase "before it exits to the left" is appropriate to a being acute, as in the figure. If you want to include a being obtuse, so that the ray would not exit to the left, that just gives you "forever" as the time frame to consider the ray's point of closest approach. Although of course you wouldn't need "forever" to determine it. In other words, the wording is not meant to be tricky: assume a to be acute if you like.

Maybe I misunderstood. You may have the answer without actually saying it. You're certainly on the right track. The question what is the distance of closest approach to O, and there is no condition that it must coincide with the incident ray when it leaves. I read your answer to depend on a. Did you mean that?

You have it for submultiples of pi radians. Can we generalize to arbitrary a?

Y-San, I researched the first version of this puzzle. It's challenging, even after reading the correct strategy. I'm very interested to see what is suggested for the current version.

By the way I am marking Rob_G's answer as the solution. Still, for a coveted bonanova Gold Star, can RG's answer be shown to be completely independent of a? It's a pretty proof.

Regular?

Wow ... talk about a "bait-and-switch." And this is such a nice analysis ... Our Puzzle Quality Control Department has informed us that we cannot claim an "aha!" status for a puzzle that admits to a straightforward constructive solution, as this one does, brilliant and insightful though it may be. They inform us that it was incorrect on our part to specify the angle a at all. Ouch. We do apologize. Can the puzzle still be answered?

On Main Street in our town stands a row of five houses, each painted a different color. They are owned by gentlemen of five different nationalities who care for five different types of pets, including a lizard, enjoy five different beverages, including vodka, and smoke four different tobacco products -- one of them is a non-smoker. From the clues below, we wonder: what beverage does the lizard owner drink, and in what color house does the vodka drinker live? The Canadian lives in the pink house. The German was bitten by his dog last week. One sometimes smells coffee near the green house. The Dane loves to drink tea. The green house is next to the beige one - to the right. The goldfish is owned by the non-smoker. The guy in the yellow house smokes "Chesterfield". In front of the middle house there are milk bottles every morning. The Norwegian lives in the house on the right. The lion keeper lives next to the "Kool" smoker, while the "Chesterfield" smoker lives next to the horse owner. The "Heart-Attack" smoker loves orange juice. The Russian smokes cigars. The blue house is next to the house of the Norwegian.

one one one three three eleven ------ twenty Solve for the letters. All are distinct. None of o, t, e are zero.

Rainman and BMAD were out for a walk the other day when they came across some barracades along a parade route that blocked their path. So much for that, mused BMAD, it looks like we either turn back or wait to watch a parade. I've got a puzzle idea, replied Rainman. Let's measure how long the parade is. When it arrives, you start walking slowly alongside it, and I'll do the same, but in the opposite direction. When I get to the end of the parade I'll stop. When the end of the parade catches up to you, you call my cell phone. BMAD agreed. When BMAD made the call, she had walked 12 blocks. Rainman reported that he had walked only a third as many. Assuming all speeds were constant, and Rainman and BMAD's speeds were the same, how long was the parade?

Ya think maybe we just did someone's homework problem?

A ray of light encounters a pair of angled mirrors from the left, as shown, a distance d from one mirror and at an angle a from the other mirror. If angle a is exactly 22.5o, how close will the ray get to their intersection at point O before it eventually exits again to the left? The drawing may not be to scale.

k-man, you're right. I took the problem as minimizing the number of Pythagorean steps, even if doing O(n2) simpler ones. I like your approach. You do programming and algorithms better than I do.

Good point. I have edited the OP by adding commas to clarify it. "A man" is the subject of all the verbs in the sentence.

A man, who was born before his father, killed his mother and then married his sister. Yet he was guilty of no wrongdoing. How is this possible?

Agree.

Some lines go off the paper and others terminate on other lines. Is it correct to assume that all lines must terminate either on the paper's edge or a previously drawn line? But if so, why is there not one line (the first one drawn) that terminates on two paper edges? Does the strategy entail the order chosen to draw the lines? Can you make clearer what is meant by "ensuring that the border defines area that is closest to the receptive point." Thanks. It sounds interesting. Edit: OK I think I get it. The points uniquely determine the lines, which are not part of the strategy. Cells that surround each point comprise the points closer to it than to any other point. The strategy comes in placing your points.

Pickett: There is still time ... k-man: thanks for being a sport - and you are well respected. But, read the ending carefully.

Sure ... if you deal me only TWO cards. But you deal me THREE cards. |----12.25----card1----12.25----card2----12.25----card3----12.25----| |----12.25----locard------------24.50------------hicard----12.25----| |----LOWER----locard-----------BETWEEN-----------hicard---HIGHER----| |--------------------------------49---------------------------------|

I don't know that it applies, although it sounds related. As I recall hearing it, it holds that a system will visit every available state given enough time. A weaker "quasi" ergodic theorem says that a system will visit a finite region around every available state (implying there is a distance function of some type) given enough time. The resolution of your conjecture may not be covered by either of these. The resolution may rest on certain conditions being satisfied. It's a zero-times-infinity puzzle that might require either a strong intuition or great mathematical skill to answer. I have some intuition, but it's often wrong. That fact keeps me coming back to this site. So I wonder whether the decimal expansion for e can occur within the expansion for pi.

At the annual cinco de mayo Brain Den party a few days ago I was joined by plasmid, k-man and Yoruichi-san, among others. I happened to mention to Y-san that the ages of my three children make for a nice puzzle. I ask for their ages after giving their sum and product, I said. Y-san said the puzzle was well known, and she seemed surprised that I would use it. It may not be that easy, I replied. In fact, at this very party four years ago I gave the challenge to k-man, along with the information that my oldest child had not yet turned 17, and he was unable to solve the problem. With as much respect as I have for k-man, Y-san replied, I don't think that means the puzzle is unsolvable. But just this evening I posed the question to plasmid, adding that my oldest child had not yet turned 21, and he too was unable to give me their ages. Well do any two of the ages differ by 1? she asked. After I answered, Y-san took my napkin and, with a wink and a smile that only she knows, wrote down the three ages. What are they?