Simple Geom problem.
 

:smile:

Here is a simple problem from the Maths Tripos.

ABC is an isoceles triangle, whose vertex angle at A is 20* . The point D is on AC such that DBC is 60*. E is on AB such that ECB =50. Find BDE

Mally :coffee:

[spoiler]30 degrees, but I am still looking for a solution that doesn't rely on trigonometry.[/spoiler]

:rolleyes:
You are correct philmoore.

This being a problem strictly on geometry , any other solution by trig, Co-ord etc is not permissble.

Hint: Dont go angle chasing. Instead try a simple construction.

Mally :coffee:

I'm pretty sure I've seen this problem before.
If the the answer is 30 degrees then if C' ia the reflection of
C in AB then C'ED is a straight line. Does this get us anywhere?

Love,
David

[quote=davieddy;98949]I'm pretty sure I've seen this problem before.
If the the answer is 30 degrees then if C' ia the reflection of
C in AB then C'ED is a straight line. Does this get us anywhere?

Love,
David[/quote]

Why don't you respond to this Mally?
It is a good problem and you posed it.
It might also keep you out of trouble meddling
in posts you don't understand.

David

[QUOTE=davieddy;99285]Why don't you respond to this Mally?
It is a good problem and you posed it.
It might also keep you out of trouble meddling
in posts you don't understand.

David[/QUOTE]

Well David I did and gave a crucial hint but somehow my post did not register.

Well here it is again.

Keeping the same notation/lettering/terminology simply mark E' on AC so that E'BC = 20*. Therefore all 3 triangles EBC , BE'C and DE'B are all isosceles.
Hence BEE' is equilateral and triangle EE'D is isosceles.
Now take it from there David and you"ll get home

Mally

Very nice problem and solution, although I don't think that the construction is at all obvious, and I would not consider it "simple"! I constructed the point F on AB such that DF and DE had the same length, hoping that I could then prove that DF was parallel to CE. I was able to prove it, but only by using the triple angle formula cos(3x) = 4cos[sup]2[/sup]x - 3cos x where x was 10 degrees, so cos(3x) was 1/2. So messy, I was really looking forward to seeing a simpler proof.

I have an interesting diagram inspired by Davieddy's comment, but I will have to wait until Monday to upload it.

[quote=mfgoode;99286]
Keeping the same notation/lettering/terminology simply mark E' on AC so that E'BC = 20*. Therefore all 3 triangles EBC , BE'C and DE'B are all isosceles.
Hence BEE' is equilateral and triangle EE'D is isosceles.
Now take it from there David and you"ll get home

Mally[/quote]

Thankyou Mally.
Even if I had seen that solution before, I'm not surprized I'd
forgotten it! As philmoore says, not that obvious or simple.

Furthermore how do you spot all the isosceles triangles if you
don't go "angle chasing" as you advised us against?

David

[quote=philmoore;99340]
I have an interesting diagram inspired by Davieddy's comment, but I will have to wait until Monday to upload it.[/quote]

Would it be a nonagon or 18-agon with some coincidences on the diagonals?

This problem suggests that choosing a multiple of 36 for the number
of degrees in a circle was inspired. When does it date from?

David

[QUOTE=philmoore;99340]Very nice problem and solution, although I don't think that the construction is at all obvious, and I would not consider it "simple"! .[/QUOTE]

Well philmoore I warned you that it is from the Maths Tripos, which is by all standards, a stiff examination of Cambridge Univ. England.

I agree its not obvious or simple. It requires, I would say, to think of a few steps ahead and look out for short cuts to the solution.

Hence I said - not to go angle chasing but rely on a construction first, and after that naturally work out the angles.

As the saying goes 'One man's meat is another man's poison'

Glad you liked it

Mally

[quote=philmoore;99340]
I have an interesting diagram inspired by Davieddy's comment, but I will have to wait until Monday to upload it.[/quote]

I still wait with interest, Phil

David

[quote=davieddy;99361]This problem suggests that choosing a multiple of 36 for the number
of degrees in a circle was inspired. When does it date from?[/quote]After I googled "Origin of 360 degrees in a circle", the first two hits,

[URL]http://mathforum.org/library/drmath/sets/select/dm_circle360.html[/URL]

and

[URL]http://www.wonderquest.com/circle.htm[/URL]

eventually both lead to the apparently authoritative book [U]The Exact Sciences in [/U]
[U]Antiquity[/U] by Otto Neugebauer.

Not having Neugebauer's reference at hand, I combined April Halladay's answer at WonderQuest with the eventual answer at MathForum, plus a little of my own, to get (in my wording):

In Mesopotamia the Sumerians had, by 2400 BC, a calendar of 12 months of 30 days each. Apparently they valued the arithmetic niceties of the number 360 more than they were irritated by the five-day yearly discrepancy. They also invented the 360-degree circle, but not subdivisions of degrees, which came later.

About 1500 BC, Egyptians invented the 24-hour day, but with variable-length hours. Roughly the same time in Mesopotamia, Babylonians invented base-60 arithmetic. Later, Greeks made the hours equal and constant.

About 300-100 BC, Babylonians subdivided both the degree and the hour into 60 minutes of 60 seconds each.

[QUOTE=davieddy;100057]I still wait with interest, Phil[/QUOTE]

I created the graphic with a program called Geometer's Sketchpad which saves its work in the form of a program-specific graphics file. I am able to copy the displayed graphic and paste it into a Word document, but I do not know how to convert it into a format (jpg or gif ?) that I can upload to the forum. Suggestions are welcomed.

[COLOR=White].[/COLOR]

That's just what I anticipated!

Thanks for posting that, Mike. I'll try to explain what I found interesting. It was provoked by David's observation of reflection. Suppose we label the vertices consecutively as V[sub]1[/sub], V[sub]2[/sub], V[sub]3[/sub], ... V[sub]18[/sub]. Then what we notice is that the following line segments are all concurrent: V[sub]1[/sub]V[sub]7[/sub], V[sub]2[/sub]V[sub]9[/sub], V[sub]3[/sub]V[sub]12[/sub], and the reflections of the first two segments across the last: V[sub]4[/sub]V[sub]15[/sub] and V[sub]5[/sub]V[sub]17[/sub]. My guess is that there is some underlying symmetry that explains why these lines are concurrent, but I haven't done much research on it. Could something related to Pascal's theorem be at work here?

Are there other regular polygons with similar
concurrencies of diagonals?

David

A diagram without the diameters might look
more spectacular?

Perhaps, but it is the fact that the intersections of the other diagonals fall directly on those diameters that is of interest.

[quote=philmoore;100126]Perhaps, but it is the fact that the intersections of the other diagonals fall directly on those diameters that is of interest.[/quote]
Two lines intersecting each other is not interesting
which is why removing the diameters (and the trivial symmetric
triple intersections) would emphasize what we are trying to illuminate.

But it is the fact that three lines intersect at one point that is interesting, and one of those lines is the diameter.

I'm sure I have a reply to this. Just give me time:)

The coincidence is that the intersection of
V(n-5) and V(n+5) lies the same distance from
the centre as that between V(n-6) and V(n+6).
That this occurs on a diagonal is unsurprizing.

My terminology needs elaboration, but I'm sure you are up to it!

[quote=davieddy;100137]My terminology needs elaboration, but I'm sure you are up to it![/quote]
On second thoughts, I think it needs much elaboration!

[QUOTE=davieddy;100138]On second thoughts, I think it needs much elaboration![/QUOTE]

Frankly your one liners are getting boring as they lack adequate info content as you yourself have said above.

A hundred thousand welcomes !

Mally

[QUOTE=cheesehead;100066]
In Mesopotamia the Sumerians had, by 2400 BC, a calendar of 12 months of 30 days each. Apparently they valued the arithmetic niceties of the number 360 more than they were irritated by the five-day yearly discrepancy. They also invented the 360-degree circle, but not subdivisions of degrees, which came later.
.[/QUOTE]

Not exactly. The Egyptians were well aware of the discrepancy and always added the 5 days at the end of their year.
The Great Pyramid was constructed over 5000 years ago (Cheops circa 4000 B.C.)
These and other secrets were encrypted in the Kufru including the accurate values of Pi , epsilon, phi. and Tau, i , 0 , infinity the seven Royal numbers.
Recommended reading "Pyramid Prophecies" by Max Toth

Mally

[QUOTE=davieddy;100136]The coincidence is that the intersection of
V(n-5) and V(n+5) lies the same distance from
the centre as that between V(n-6) and V(n+6).
That this occurs on a diagonal is unsurprizing.[/QUOTE]

If you mean that it occurs on a diameter is unsurprising, then what you say is correct. Returning to my earlier notation, we have the diameter V[sub]3[/sub]V[sub]12[/sub], and the two pairs V[sub]2[/sub]V[sub]9[/sub] and V[sub]4[/sub]V[sub]15[/sub], and also V[sub]1[/sub]V[sub]7[/sub] and V[sub]5[/sub]V[sub]17[/sub]. We can either prove that the diameter and a single segment from each pair are concurrent, or we can prove that one segment from either pair is concurrent with both members of the other pair.

Yep.

[quote=mfgoode;100150]Frankly your one liners are getting boring as they lack adequate info content as you yourself have said above.
Mally[/quote]

I think your penchant for verbosity is widely understood :)

[quote=philmoore;100095]Thanks for posting that, Mike. I'll try to explain what I found interesting. It was provoked by David's observation of reflection. Suppose we label the vertices consecutively as V[sub]1[/sub], V[sub]2[/sub], V[sub]3[/sub], ... V[sub]18[/sub]. Then what we notice is that the following line segments are all concurrent: V[sub]1[/sub]V[sub]7[/sub], V[sub]2[/sub]V[sub]9[/sub], V[sub]3[/sub]V[sub]12[/sub], and the reflections of the first two segments across the last: V[sub]4[/sub]V[sub]15[/sub] and V[sub]5[/sub]V[sub]17[/sub]. My guess is that there is some underlying symmetry that explains why these lines are concurrent, but I haven't done much research on it. Could something related to Pascal's theorem be at work here?[/quote]

With your notation, Mally's constuction line is V[sub]2[/sub]V[sub]5[/sub].
This must help somehow.

Remind me of Pascal's theorem please.

David

360 degrees
 

Thanks for the history lessons.
Approx 360 days in a year was (I would hope) an
unimportant distraction. But a hexagon consisting
of 6 equilateral triangles allied to base 60 arithmetic
seems eminently sensible.

David

Base 60
 

How did the Babylonians (or whoever) represent their 60 digits?

[quote]How did the Babylonians (or whoever) represent their 60 digits?[/quote]
We're assuming this is not a rhetorical question:

[url]http://en.wikipedia.org/wiki/Babylonian_number_system[/url]

[QUOTE=davieddy;100222]How did the Babylonians (or whoever) represent their 60 digits?[/QUOTE]They used a mixture of base 10 and base 60, alternating both :
1, 10, 60, 600 (10*60), 3 600 (60*60), 36 000 (10*60*60)...
A symbol for the units, another symbol for ten, another for 60, superposition of the previous two for 600, another for 3 600, superposition of the symbol ten and 3 6000 for 36 000...

For instance u representig the unit, t the ten , s the sixty, r the 600, and o the 3 600, p the 36 000 : pppp ooooo rrrrr ss ttttt u would be 36000+36000+36000+36000+3600+3600+3600+3600+3600+600+600+600+600+60+60+10+10+10+10+10+1=164571
The symbols where grouped in paterns to facilitate reading.
(source Histoire universelle des chiffres by George Ifrah, editor Seghers 1981)

Xyzzy was quicker to answer :-)

[quote=Xyzzy;100223]We're assuming this is not a rhetorical question:

[URL]http://en.wikipedia.org/wiki/Babylonian_number_system[/URL][/quote]

It wasn't. Remembering ~26 letters of the alphabet seems
to be a sensible limit.
BTW was that the royal "we"?

David

[quote=davieddy;100225]BTW was that the royal "we"?[/quote]No. Xyzzy uses the [I]editorial we[/I] (he being head honcho hereabouts), or perhaps a generic third person (See the "Atypical uses of we" section at [URL]http://en.wikipedia.org/wiki/We[/URL]), or [I]pluralis modestiae[/I] (rather than [I]pluralis majestatis[/I], the [I]royal we[/I]).

OTOH, Xyzzy does have a tendency to do things like inserting false quotations (e.g. [I]"I quote more ... all day!"[/I]) between one's nym and avatar in the left margin, which seem more imperial than editorial.

[quote=cheesehead;100238]OTOH, Xyzzy does have a tendency to do things like inserting false quotations (e.g. [I]"I quote more ... all day!"[/I]) between one's nym and avatar in the left margin, which seem more imperial than editorial.[/quote]

Speaking of which, what does mine mean?

[quote]Speaking of which, what does mine mean?[/quote]

:google:

BTW, for the record, the actions that we take (Xyzzy, and his evil twin Yzzyx) are, for the most part, random, malicious and devoid of meaning.

Some misguided souls have hypothesized that there is a relationship between various psychotropic medications and our behavior but we dismiss that line of thought as a desperate attempt to make something utterly pathetic and unfathomable into something sensible. Sometimes things really stink, no matter how much perfume you put on them.

:mally:

[quote=cheesehead;100238]Xyzzy uses the [I]editorial we[/I] (he being head honcho hereabouts), or perhaps a generic third person (See the "Atypical uses of we" section at [URL]http://en.wikipedia.org/wiki/We[/URL]), or [I]pluralis modestiae[/I] (rather than [I]pluralis majestatis[/I], the [I]royal we[/I]).[/quote]
[quote=Xyzzy;100287]BTW, for the record, the actions that we take (Xyzzy, and his evil twin Yzzyx)[/quote]Well, now that "Yzzyx" is on record, I wish to amend my previous statement.

Xyzzy's earlier use of the [I]editorial we [/I]has now progressed into the [I]delusional we[/I]. That is, his [I]use[/I] has progressed; Xyzzy himself is not actually delusional, but seems desirous of presenting that facade.

Irish.
 

[QUOTE=davieddy;100252]Speaking of which, what does mine mean?[/QUOTE]

Its Irish for "a thousand welcomes"

I thought you would get it in my previous post to you when thats how I signed off but you were too busy with my verbosity.

Mally

Cracked it!
 

[quote=mfgoode;97918]:smile:

Here is a simple problem from the Maths Tripos.

ABC is an isoceles triangle, whose vertex angle at A is 20* . The point D is on AC such that DBC is 60*. E is on AB such that ECB =50. Find BDE

Mally :coffee:[/quote]

Three years ago, this puzzle was entertaining philmoore and me.
I have revisited it and come up with this solution:

[spoiler]Let F be the reflection of B in AC and
let G be the reflection of C in AF.
AD=DB and AGB is equilateral
Let DG intersect AF at H
DGF is 50 degrees so AH=AE

BDE=FDG=30[/spoiler]

David

.

Hint to graphics fans
 

(Myself included)

A picture must be worth at least 50 words:smile:

David

PS Or the other way round
(as the actress said to the bishop)

Can't say I follow your solution yet.
[SPOILER]Why is angle DGF 50 degrees?[/SPOILER]

[quote=philmoore;220028]Can't say I follow your solution yet.
[spoiler]Why is angle DGF 50 degrees?[/spoiler][/quote]
[spoiler]BGF = 20 (e.g. because BAF is 40 or AGB is 60) and DGB is 60/2[/spoiler]

David

[quote=Xyzzy;100082][COLOR=white].[/COLOR][/quote]

Click this post for the diagram.

[quote=philmoore;100095]Thanks for posting that, Mike. I'll try to explain what I found interesting. It was provoked by David's observation of reflection. Suppose we label the vertices consecutively as V[sub]1[/sub], V[sub]2[/sub], V[sub]3[/sub], ... V[sub]18[/sub]. Then what we notice is that the following line segments are all concurrent: V[sub]1[/sub]V[sub]7[/sub], V[sub]2[/sub]V[sub]9[/sub], V[sub]3[/sub]V[sub]12[/sub], and the reflections of the first two segments across the last: V[sub]4[/sub]V[sub]15[/sub] and V[sub]5[/sub]V[sub]17[/sub]. My guess is that there is some underlying symmetry that explains why these lines are concurrent, but I haven't done much research on it. Could something related to Pascal's theorem be at work here?[/quote]

When I looked at your diagram again, all became blindingly
obvious (which is why my "proof" above was a bit casual).
Let us call the centre A and the "multiple coincidences" v[sub]n[/sub].
I spotted that V[sub]-3[/sub]v[sub]-1[/sub]v[sub]1[/sub]V[sub]3[/sub] bisected AV[sub]0[/sub] perpendicularly.
The reason is obvious: V[sub]-3[/sub]V[sub]3[/sub]V[sub]9[/sub] is an equilateral triangle.

Talk about a penny dropping!

David

Note also that just as AV[sub]1[/sub] and V[sub]0[/sub]V[sub]7[/sub]
intersect on V[sub]-3[/sub]V[sub]3[/sub] (D in the original problem),
so do AV[sub]2[/sub] and V[sub]0[/sub]V[sub]5[/sub] (e.g. H or E)

David

Regular 18-agon
 

[quote=philmoore;220028]Can't say I follow your solution yet.
[spoiler]Why is angle DGF 50 degrees?[/spoiler][/quote]

I'm sure that you intended this as a friendly, tactful hint from
one busy teacher to another (less busy) that I was a bit slapdash.
I don't believe you couldn't work out the angle between
two diagonals of a regular 18-agon!

OK let's start with that.
For our purposes, sides are diagonals.

V[sub]0[/sub]V[sub]1[/sub] is // to V[sub]-1[/sub]V[sub]2[/sub] etc up to V[sub]-8[/sub]V[sub]9[/sub].
(Clock arithmetic of course).
Rotating through 20 degrees 8 times gives us 9*9 diagonals.

V[sub]-1[/sub]V[sub]1[/sub] is // to V[sub]-2[/sub]V[sub]2[/sub] etc up to V[sub]-8[/sub]V[sub]8[/sub].
Rotating through 20 degrees 8 times gives us 9*8 diagonals.

9*9 + 9*8 = 18C2

V[sub]0[/sub]V[sub]1[/sub]V[sub]-1[/sub] = 10 degrees.

[U]Now for my solution to the original problem:[/U]

Let the centre of the 18-agon be A. Let V[sub]0[/sub] be B and V[sub]1[/sub] be C.
V[sub]-3[/sub]V[sub]3[/sub] bisects AB perpendicularly.
(AV[sub]+/-3[/sub]B are equilateral).

V[sub]1[/sub]V[sub]0[/sub]V[sub]7[/sub] = 60
AV[sub]0[/sub]V[sub]7[/sub] = 20
V[sub]0[/sub]AV[sub]1[/sub] = 20
V[sub]0[/sub]V[sub]7[/sub] intersects AV[sub]1[/sub] on V[sub]-3[/sub]V[sub]3[/sub] at D.
Reflect V[sub]-3[/sub]V[sub]3[/sub] in AV[sub]1[/sub] to get V[sub]-1[/sub]V[sub]5[/sub] which passes through D. It intersects AV[sub]0[/sub]
at E?
Reflect V[sub]-1[/sub]V[sub]5[/sub] in AV[sub]0[/sub] to get V[sub]1[/sub]V[sub]-5[/sub].

Show that V[sub]-5[/sub]V[sub]1[/sub]V[sub]0[/sub] = 50 degrees,
confirming that E? = E.

Now find V[sub]0[/sub]DV[sub]-1[/sub]

David