wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> The Peaks of Berkeley
        
(Message started by: rloginunix on Apr 4^th, 2015, 1:20pm)

Title: The Peaks of Berkeley
Post by rloginunix on Apr 4^th, 2015, 1:20pm

The Peaks of Berkeley.

One fine morning Willy Wu was walking with constant velocity to the Euclidean Geometry lecture when exactly at 09:00 am he saw the top of the Sather Tower under the angle of 45 degrees. At 09:05 am the angle was 60 degrees. When Willy passed the tower, at 09:15 am, the angle was http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif degrees:

http://www.ocf.berkeley.edu/~wwu/YaBBAttachments/rlu_tpob01.png

Find http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif.

How would the answer change if the second time, when Willy measures the 60-degree angle, he already passed the tower:

http://www.ocf.berkeley.edu/~wwu/YaBBAttachments/rlu_tpob02.png

One type of solution is trivial. If you will be submitting it keep in mind that the answer is a perfectly round number - please obtain it without the calculator.

The reason this puzzle is in the medium section, however, is because it is a bit more challenging to find the other type of solution(s). I have found two, you may do better.

(Assume that Willy's path and the Sather Tower are in the same plane)

Title: Re: The Peaks of Berkeley
Post by jollytall on Apr 5^th, 2015, 12:12am

If the tower is 1 high, then the starting point from the bottom of the tower at 9:00 is tg 45 = 1. At 9:05 he is tg 30 = 1/sqrt(3). So his speed is (1-1/sqrt(3)) / 5 minutes. He is walking for 15 minutes, i.e. 3-sqrt(3) tower height, i.e. he is 2-sqrt(3) from the tower.
Knowing that tg (a+b) = (tg a + tg b)/(1-tg a * tg b), it so happens that tg (45-30) = (1-1/sqrt(3))/(1+1*1/sqrt(3))=(sqrt(3)-1)/(sqrt(3)+1) =2 - sqrt(3).
This is exactly how far he is from the tower at 9:15, i.e. his angle from the top of the tower is 15; or he sees the tower top at 75 degree.

In the second case his speed is (1+1/sqrt(3))/5 minutes, his distance from the tower at 9:15 is 2+sqrt(3).
With the same tg equation
tg (45+30) = 2+sqrt(3), i.e. he is seen from the top of the tower at 75 degree, or he sees the the tower at 15 degree.

P.S. Are there two open doors at the bottom of the tower? Otherwise the answer to both questions is 90 degrees with ambulance people around.

Title: Re: The Peaks of Berkeley
Post by rloginunix on Apr 5^th, 2015, 10:55am

Yes, well done.

It is also possible to put AB = d, BC = 2d, AC = 3d and so on.

Now, forget the trigonometry and give simplicity a shot. One solution pops into view right away when you are trying to construct the required drawing preparing it for a post - how does one ensure the above observation?

To intrigue the audience a little bit there is combinatorics lurking beneath this with related articles from Cornell and the Electronic Journal of Combinatorics.

Also, do not sweat the construction - if someone supplies the words I will supply the drawings.

PS
Not sure about the doors at the bottom of the Sather Tower - 've been to [hide]Princeton, Harvard, MIT[/hide] campuses but never made it to the West coast (yet). Anyway, Willy Wu possesses the properties of a neutrino - he can pass through things undamaged ...