Here is a fun one: The triangle ABC is isoseceles, with *angle*(ABC) = *angle*(ACB) = 80°. We draw now two segments inside the triangle. First, we choose the point D on segment AC such that *angle*(CBD) is 60°. Also, we choose the point E on AB, where *angle*(BCE) is 50°. We now extend DE until it intersects with the line on the BC segment in point F, as below:

The question is: what is the angle between the two lines on F? In other words, what is the value of *angle*(DFC)?

30 degrees.

30 deg

Would someone post how they got the solution at some point? I, with my hate of geometry, plugged at this problem and came up short. But I’d be anxious to see how the solution was gotten.

Though, to be fair to other people who want to try on it before being given the solution, perhaps post the solution on another page and link to it, or obfuscate it in some other fashion.

BAC = 20

EBD = 20

ECD = 30

EGD = 70 (G being the middle point that isn’t named)

BGC = 70

EGB = 110

DGC = 110

ADB = 140

AEC = 130

Therefore:

FDB + DFB + 120 = 180

FDB + DFB = 60

You can take any two values adding up to 60 for both angles and plug them in and all triangles still add to 180 as do all straight lines.

What am I missing?

Because without the length of at least one side you cannot find the angle because FB and FD could be any length and we don’t have enough information (i.e. we have two variables that changing either and keeping the sum will work) to extrapolate DFB to an absolute value.

Hence unless I screwed up (I Probably did) you can’t solve this unless we have at least one length.

Using the angles you have already worked out you now have one more isoceles triangle, can you see where?

This means that to all intents and purposes you DO have a new length to play with (albeit a length relative to other known lengths).

You can now use the full machinery of triangle geometry involving length! This means you can use the sine rule, similarity of triangles, Heron’s formula (!!)…

Some ways of finishing the problem off will be far more elegant than others, but for all intents and purposes noticing this reduces the problem to ‘just’ a computation.

Slightly offtopic:

For a really interesting tour of triangle geometry check out Paul Yiu’s introduction here:

http://www.math.fau.edu/yiu/TourOfTriangleGeometry/MAAFlorida37040428.pdf

Also see Clark Kimberling’s phenomenal compilation of triangle centers:

http://faculty.evansville.edu/ck6/encyclopedia/ETC.html

hahaha!

i forwarded this around to my office, and everyone is stuck working on it….at first it irritated me that i couldnt solve it, but now i take pleasure in knowing that the rest of them are struggling too…

😛

You are an evil, evil man, Adi. Valorie and I have been on this WAY too long

BAC = 20

ABD = 20

BEC = 50

BED = 70

EDB = 40

DEB = 70

AED = 60

ADE = 100

BEF = 60 (since FED = 180)

so eventually got

BFE = 20

oops

meant to say CED = 70 (not BED); ok is this correct ( I will give all the calculations)

AED = 60

ADE = 100

EDB = 40

DEC = 70

BDC = 40

BEC = 50

FBE = 100 since B on the line FBC = 180

FEB = 60

so

BFE = 20

The imaginary spot G (where lines criss-cross)

EGD = 70

EGB = 110

CGD = 110

BGC = 70

Is it cheating to use CAD software? Anyway, AutoCAD verified the 30 degrees 😉

I can get

this, it will take some time…

At work you feel rushed to finish…

…

spent an hour..seems like you’re almost there but not there yet 🙁 doesn’t seem to be too difficult but tricky enough to eat your brains. just hope its not something too trivial :-S

The algebra is complicated enough that I don’t want to type it in. However, through a combination of the Law of Sines and the fact that, for any angle, theta, such that theta < 180 degrees (i.e. theta is any interior angle of a triangle), sin(theta) == sin(180-theta), you can show that triangle ADE is similar to the triangle formed by points B, E and the intersection of line segments BD and CE. (Hint: sin(BCE) == sin(AEC); i.e. AEC + BCE == 180 degrees.)

From the similarity of these two triangles, we deduce that angle AED is 50 degrees, and its opposite angle, BEF, is also 50 degrees. Since FBE is 100 degrees (adjacent to angle CBE = 80 degrees), angle BFE must be 30 degrees.

Search on Google for "isosceles 80" and click on the first result.

Argh. Ok, so I couldn’t just use my two trivial equations of triangle angle and straight lines 🙂

Oh well! Thanks 🙂 (I was hoping that SOCATOA didn’t come into it, I always hated that stuff!)

40/3 deg

Eight solutions are presented here:

http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf

One of them even (#4) in flash format:

http://agutie.homestead.com/files/LangleyProblem.swf

Surprisingly, the solution that I know is not among them – I’ll post it later after I complete the drawing…

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