Geometry Stumper

[Albert Chan]

I got the interesting problem from book Return of the Straight Dope

Solution is on the web (link below), but try not to peek.

Can Cecil solve this geometry stumper ?

Quote:Dear Cecil:

The following geometry problem has haunted me for 30 years. Every math teacher,
Ph.D., engineer, etc., I've shown it to was stumped. Now it's your turn.

Given any triangle ABC, extend the baseline and bisect the interior and exterior angles at A and B.
Extend these four new lines until they intersect at D and E. Prove DCE is a straight line.
(It looks straight, but no one can prove it.) How about it?

— John Ricketts, Madison, Wisconsin

Code:

 D
                 C
                                   E
             
            
         
 F       A ------------------ B    Z
             Triangle base

Point D:
Angle(FAC) = 2 * Angle(FAD)
Angle(ABC) = 2 * Angle(ABD)

Point E:
Angle(ZBC) = 2 * Angle(ZBE)
Angle(BAC) = 2 * Angle(BAE)

Prove: DCE is a straight line

[Thomas Puettmann]

In a university course on elementary geometry you gradually build up skills to solve these problems almost immediately.

Some things just flash when you read the problem.

Angular bisectors --> inscribed circle --> buy two, get one free

Now the only point is that the circle in question is not the inscribed circle of the triangle ABC, but the circle whose midpoint is D and which touches the three lines AB, AC, and BC.

Maybe some of you would like to give this approach a try.

[Albert Chan]

(08-14-2018 06:04 PM)Thomas Puettmann Wrote:  the circle in question is not the inscribed circle of the triangle ABC, but
the circle whose midpoint is D and which touches the three lines AB, AC, and BC..

What does midpoint of a circle mean ?

Is it true that a circle that touches AB, AC, BC must be inscribed or circumscribed ?
My geometry is a bit rusty ...

[Thomas Puettmann]

(08-14-2018 06:58 PM)Albert Chan Wrote:  What does midpoint of a circle mean ?

center, sorry.

A standard theorem from elementary geometry says that all three (interior) angular bisectors meet at one point, namely, the center of the inscribed circle.

How can one proof this? Imagine an arbitrary point on an angular bisector. Keeping this point as the center start blowing this point into a larger and larger circle. For one and only one radius the circle will touch both rays of the angle simultaneously.

Now, imagine you have two (interior) angular bisectors in a triangle. They meet at one and only one point. The circle just described now touches all three sides of the triangle. This circle is called inscribed circle of the triangle. It follows immediately that the third angular bisector must pass through the center of this circle.


To conclude the proof of your original problem you do the same at D outside of the triangle ABC. Just watch out what you get for free.

To those who like problems like this and would like to develop some geometric reasoning skills, I cannot recommend highly enough the book "Modern Geometry with Applications" by George A. Jennings. This is the book I really learned geometry from (actually quite late when I had to give my first course on the subject), and it is still one of my favourite books.

[Albert Chan]

(08-14-2018 08:35 PM)Thomas Puettmann Wrote:  A standard theorem from elementary geometry says that all three (interior) angular bisectors meet at one point, namely, the center of the inscribed circle.

I "see" the solution !

The twist in this puzzle is the "inscribed" circle is on the outside.
But the circle still "inscribed" the lines, if you extend the lines.

If you extend line BC, and you are at B, there is 3 angles left of C
Angle(ACB) + 2 * Angle(ACD) = 180 degree (since it is just line extension)

using symmetry, the other side have the same constraint:
Angle(ACB) + 2 * Angle(BCE) = 180 degree

Add the two equations, then divide by 2:
Angle(ACD) + Angle(ACB) + Angle(BCE) = 180 degree

--> Angle(DCE) = 180 degree

Thanks.

[Voldemar]

(08-15-2018 12:39 AM)Albert Chan Wrote:  Angle(ACB) + 2 * Angle(ACD) = 180 degree (since it is just line extension)

How can You know that there are two equal angles (2*ACD). It must be proved. Known is only that ACB + ACD + unknown angle = 180°.

[Thomas Puettmann]

(08-15-2018 12:39 AM)Albert Chan Wrote:  I "see" the solution !

Great! "Seeing" is the great fun in geometry, and it is almost entirely training.

Attached is what I have sketched after I read your original post.

The original triangle is in grey, the bisectors given in the problem are in red and blue and what you get for free is in green. That is, line DC is the bisector of the exterior angles at C (which is perpendicular to the bisector of the interior angle because the exterior and interior angle sum up to 180°).

[Albert Chan]

(08-15-2018 06:38 AM)Voldemar Wrote:  

(08-15-2018 12:39 AM)Albert Chan Wrote:  Angle(ACB) + 2 * Angle(ACD) = 180 degree (since it is just line extension)

How can You know that there are two equal angles (2*ACD). It must be proved.

Circle, with center D, inscribed the lines AB, AC, BC.
Connect D with those 3 touched line, so DF, DG, DH are radius of circle

What I "see" is triangle DCG, if flipped up, you get DCH.

Code:

           H 


D
                  C


           G

F      A                   B

Prove:
Right triangle DCG and DCH are congruent, since DC=DC (same line), DG=DH (both are radius).

-> Angle(DCG) = Angle(DCH)

Angle(ACB) + Angle(DCG) + Angle(DCH) = 180 degree

-> Angle(ACB) + 2 * Angle(ACD) = 180 degree

[Voldemar]

(08-15-2018 11:14 AM)Albert Chan Wrote:  Circle, with center D, inscribed the lines AB, AC, BC.

Aha. But do not forget to prove why circle with center D touch line AC.
Because AD is the bisector of angle CAF.
Your solution is nice.

[Albert Chan]

(08-15-2018 11:56 AM)Voldemar Wrote:  Aha. But do not forget to prove why circle with center D touch line AC.
Because AD is the bisector of angle CAF.
Your solution is nice.

Good catch.

Technically, you have to proof that circle really inscribe the lines from the outside.
(like you did with why circle touch AC, but need 2 more to complete proof)

Cecil's prove only uses triangles. (see my first post for link)

He likely know about the circle, but it required more explanations to readers in his column.
Instead, he proved backwards (not using circle), showing DF = DG = DH

BTW, have you peek at Thomas Puettmann sketch, about the green lines ?
Why is that a right triangle ?

[Voldemar]

(08-15-2018 12:29 PM)Albert Chan Wrote:  BTW, have you peek at Thomas Puettmann sketch, about the green lines ?
Why is that a right triangle ?

There are 4 angles at C. Two angles x and two angles y.
2x + 2y = 180
Divide by 2
x + y = 180/2 = 90
Angle x + y is our angle in the green lines.

[Albert Chan]

Here is the prove, using power of Mathematica

Triangle scaled so base = 1, height = 1
Setup so A = {0,0}, B = {1, 0}, C = {k, 1}

If CDE is straight line with scaled triangle, it will still be straight, unscaled

Code:

In[1]:= pointE = {x, y} /. Solve[y == ae*x == be*(x - 1), {x, y}] // First

             be       ae be
Out[1]= {-(-------), --------}
           ae - be   -ae + be

In[2]:= pointD = pointE /. {ae -> -1/ae, be -> -1/be} // Simplify

           ae        1
Out[2]= {-------, --------}
         ae - be  -ae + be

In[3]:= pointC = {k, 1};

In[4]:= bisect = t /. Solve[{1/k == 2t/(1 - t*t)}, t] (* solve for ad, ae *)

                        2                  2
Out[4]= {-k - Sqrt[1 + k ], -k + Sqrt[1 + k ]}

In[5]:= ae = Last[bisect] (* pick the positive slope *)

                       2
Out[5]= -k + Sqrt[1 + k ]

In[6]:= be = ae /. k -> k-1 (* line BC slope = 1/(k-1), not 1/k *)

                             2
Out[6]= 1 + Sqrt[1 + (-1 + k) ] - k

In[7]:= Det[{pointD - pointC, pointE - pointC}] == 0 // Simplify (* CDE on the same line ? *)

Out[7]= True

[Albert Chan]

There is an even simpler Mathematica prove, without finding out slopes

Code:

In[1]:= pointE = {x, y} /. Solve[y == ae*x == be*(x - 1), {x, y}] // First;
In[2]:= pointD = pointE /. {ae -> -1/ae, be -> -1/be} // Simplify;
In[3]:= pointC = {k, 1};
In[4]:= Det[{pointD - pointC, pointE - pointC}] // Together
           2          2        2                   2             2
        -ae  - be - ae  be + be  - ae k + be k + ae  be k - ae be  k
Out[4]= ------------------------------------------------------------
                                          2
                                 (ae - be)
In[5] = % /. {ae^2 -> 1-2*k*ae, be^2 -> 1-2*(k-1)*be} // Expand           
Out[5]:= 0

Above only used the half-angle formula: tan(2x) = 2 tan(x) / (1 - tan(x)^2)

tan(Angle CAB) = 1/(k-0) = 2*ae / (1 - ae^2), thus ae^2 = 1-2*k*ae
tan(Angle CBA) = 1/(k-1) = 2*be / (1 - be^2), thus be^2 = 1-2*(k-1)*be

since determinant = 0, slope DC = slope CE, thus DCE is straight line

[Albert Chan]

(08-15-2018 01:16 PM)Voldemar Wrote:  

(08-15-2018 12:29 PM)Albert Chan Wrote:  BTW, have you peek at Thomas Puettmann sketch, about the green lines ?
Why is that a right triangle ?

There are 4 angles at C. Two angles x and two angles y.
2x + 2y = 180
Divide by 2
x + y = 180/2 = 90
Angle x + y is our angle in the green lines.

this is enough to prove the whole thing !

the other side will have two angle y and two angle z, y + z = 90

--> angle DCE = x + y + y + z = (x+y) + (y+z) = 90 + 90 = 180

Wow, the power of geometry.

[Thomas Puettmann]

You do not even need to consider the right angle at C. In my first sketch, I drew it, but it is not neccessary for the argument.

Once you know that the line DC is the angular bisector of the exterior angles at C, you are done (by symmetry, the line EC is the same angular bisector).

Buy two, get one free, thanks to the circle at D.

[Albert Chan]

(08-15-2018 09:46 PM)Thomas Puettmann Wrote:  Once you know that the line DC is the angular bisector of the exterior angles at C, you are done
(by symmetry, the line EC is the same angular bisector).

That reminds me of a book I read, "The Man Who Love Only Numbers".

Erdos once said, "You don't have to believe in God, but you should believe in the Book"

Your proof should belong in the Book.

Volemar already proved the green right angle, and is so close to the proof.
Just say, "by symmetry, the other side also got a right angle", and the proof is done.

Too bad ...

[jwhsu]

(08-14-2018 08:35 PM)Thomas Puettmann Wrote:  

(08-14-2018 06:58 PM)Albert Chan Wrote:  What does midpoint of a circle mean ?

center, sorry.

A standard theorem from elementary geometry says that all three (interior) angular bisectors meet at one point, namely, the center of the inscribed circle.

How can one proof this? Imagine an arbitrary point on an angular bisector. Keeping this point as the center start blowing this point into a larger and larger circle. For one and only one radius the circle will touch both rays of the angle simultaneously.

Now, imagine you have two (interior) angular bisectors in a triangle. They meet at one and only one point. The circle just described now touches all three sides of the triangle. This circle is called inscribed circle of the triangle. It follows immediately that the third angular bisector must pass through the center of this circle.


To conclude the proof of your original problem you do the same at D outside of the triangle ABC. Just watch out what you get for free.

Is this basically using an excircle to solve the problem? (I had to look it up, it has been a long while since I took geometry). From what I recall, one of its properties is the center of an excircle of a side of a triangle is the intersection of the two external angle biscectors and the extension of the opposite interior angle bisector of a triangle.

For this problem, if you extend sides AC and BC, the angle opposite ACB is identical. This means the angles adjacent to ACB are also identical. Using the properties of excircles, point D is the center of an excircle for line AC and segment CD bisects the two exernal angles of the triangle on that side. Similar logic goes for line BC and its corresponding excircle. This means the angles around point C are 2*ACB + 2*ACD + 2* BCE = 360°. Dividing in half gives us ACB + ACD + BCE = 180°, which means that segment DCE is a straight line.

[Thomas Puettmann]

(08-16-2018 12:30 AM)jwhsu Wrote:  Is this basically using an excircle to solve the problem?

Yes, I forgot about the name. In mathematics you try to memorize as little as possible. Here, this means essentially just the reflex "angular bisectors" --> "inscribed circle". The definition, the construction and the properties of the three excircles are so analogous to that of the inscribed circle, that you simply recover them when needed.

(08-15-2018 11:02 PM)Albert Chan Wrote:  Your proof should belong in the Book.

Euclid and his ancestors have shown how to use inscribed circles and excircles and their proofs definitely belong in the Book. "Seeing" in elementary geometry is a lot of fun, but there is almost no way to do something original. It's an art of recreation. All simple problems like this have been solved thousands if not millions of times with minor variations.

With the notion of an excircle, here goes the complete proof.

- The angular bisector of the interior angle at B and the angular bisector of the exterior angle at A meet at the center D of one of the three excircles of the triangle ABC.
- This excircle touches the three lines AB, AC, and BC.
- The line DC is the angular bisector of the exterior angle at C.

- By the symmetric construction on the other side, the line EC is also the angular bisector of the exterior angle(s) at C.
- Since the angular bisector of the exterior angles at C is unique, we have that the line DC is equal to the line EC.

No need to produce an 180° angle at C.

[Thomas Puettmann]

(08-15-2018 07:23 PM)Albert Chan Wrote:  There is an even simpler Mathematica prove, without finding out slopes

I like such CAS proofs a lot, although I haven't done any by myself. Doron Zeilberger from Rutgers has written a book, in which he proves 55 theorems from plane geometry using Maple:
http://sites.math.rutgers.edu/~zeilberg/GT.html
He has very strong opinions about that computer math is the "real" math and all what we humans still do ourselves is unimportant because it is of neglegtible complexity.

[Albert Chan]

Hi, Thomas Puettmann,

I don't know about computers replacing human to do "math".
But, it allowed you to do things you would never dare checking.

For example, how about proving a straight line by adding up distance ?
Warning: do not do this by hand ! Even CAS have a tough time ...

In[1]:= pointE = {x, y} /. Solve[y == ae*x == be*(x - 1), {x, y}] // First;
In[2]:= pointD = pointE /. {ae -> -1/ae, be -> -1/be} // Simplify;
In[3]:= pointC = {k, 1};

Code:

In[4]:= distance[pt_] := Sqrt[pt[[1]]^2 + pt[[2]]^2];
In[5]:= Solve[distance[pointD - pointC] + distance[pointC - pointE] == distance[pointD - pointE], k]
                  2          2        2
                ae  + be + ae  be - be
Out[5]= {{k -> --------------------------}}
                            2           2
               -ae + be + ae  be - ae be

In[6]:= % /. {ae^2 -> 1 - 2*k*ae, be^2 -> 1 - 2*(k-1)*be} // Simplify
Out[6]= {{k -> k}}

Solving for value of k so DCE is straight, we got {k -> k}
That means DCE is always straight