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        riddles >> easy >> Triangle with angle split into 4 equal parts
        
(Message started by: jollytall on Mar 28^th, 2014, 11:26pm)

Title: Triangle with angle split into 4 equal parts
Post by jollytall on Mar 28^th, 2014, 11:26pm

There is a triangle. From one vertex we draw the height line, the angle bisector and the median line. These three split the angle into four equal parts.
What are the angles of the triangle?

Title: Re: Triangle with angle split into 4 equal parts
Post by rloginunix on Mar 29^th, 2014, 8:09am

[hide]22.5, 67,5, 90.[/hide]

I used [hide]IV.5[/hide] and [hide]III.31[/hide] to do this.

Need to eat my breakfast first, then I'll do the GeoGebra construction and post my line of reasoning.

Title: Re: Triangle with angle split into 4 equal parts
Post by rloginunix on Mar 29^th, 2014, 5:10pm

A warning to a newcomer - if you want to solve this problem on your own please do not read any further. Full solutions are coming up (please let me know if I need to hide this).

In this and following constructions point H marks the given height, point S marks the given bisector, point M marks the given median. Angles beta are the four equal angles given.

Solution 1, A Happy Medium.

http://www.ocf.berkeley.edu/~wwu/YaBBAttachments/rlu_4ea1.png

1). From Euclid's Elements Book 4 proposition 5 (B4.P5 onward) we know that we can circumscribe a circle about a given triangle. Do that.

2). Extend the line BM until it intersects the circumcircle at point D.

3). Construct the line segment DC.

4). From B3.P21 we know that in a circle the angles in the same segment equal one another. BC is a common segment shared by the angles BAC and BDC by our construction. Hence, they are equal.

5). In the triangle ABH the angle ABH is 3*beta. In the triangle CDB the angle DBC is 3*beta. Hence, these are equal.

6). From B1.P32 (the sum of the three interior angles of the triangle equals two right angles) for the triangle ABH we have:

alpha + 3*beta + 90 = 180

For the triangle BDC we have:

alpha + 3*beta + DCB = 180

From these it follows that the angle DCB = 90 degrees.

7). But according to B3.P31 in a circle the angle in the semicircle is right and hence point M must be the circumcircle's center and DB is one of its diameters.

7+1). From that (M belongs to two line segments) and from the definition of the median (AM = MC) it follows that AC is also the circumcircle's diameter.

9). And from that it follows that the angle ABC must be 90 degrees. So the angle B = 90 degrees.

10). From 7) it follows that AM = MB, which means that triangle AMB is isosceles. From B1.P5 it follows that the angles at the base AB are equal: alpha = beta = 90 / 4 = 22.5. So the angle A = 22.5 degrees.

11). From B1.P32 it follows that:

22.5 + 90 + x = 180
x = 67.5

So the angle C = 67.5 degrees.

[edit]
Correction: fixed a typo in 6). "... DCB = 190" became "... DCB = 180".
Moved the drawing file to this forum.
[/edit]

Title: Re: Triangle with angle split into 4 equal parts
Post by rloginunix on Mar 29^th, 2014, 5:35pm

Solution 2, Bisector's Revenge.

http://www.ocf.berkeley.edu/~wwu/YaBBAttachments/rlu_4ea2.png

1). Extend the bisector BS until in intersects the circumcircle at D.

2). Construct a perpendicular to AC through D. By definition of the bisector angles ABD and DBC are equal. In reverse of B3.P21 I would say that equal angles subtend equal segments and AD = DC. Hence, the triangle ADC is isosceles and the height from D is also a bisector and a median which falls on M which sits on a common base AC. And by our construction DM is parallel to BH.

3). From 2) it follows that the angles D and DBH are vertical and from B1.P15 it follows that they are equal and angle D = beta.

4). I omit this proof but it can be proven that if the angles at the base are equal then the triangle must be isosceles. So since angle D = beta = DBM then triangle DBM is isosceles and MD must be equal MB (MD = MB).

5*). By definition of the circle it follows that OD = OB, hence the triangle DOB is isosceles and from B1.P5 it follows that the angles at the base DB are equal. Since angle D = beta then angle OBD = beta.

6). We now have two (isosceles) triangles DOB and DMB with a common base DB and equal angles at that base. According to B1.P26 these triangles are equal which means that points M and O are one and the same (in the drawing I separated them on purpose).

7). It follows then that AM = AO = OC = MC and the rest of the solution is the same as the first one - triangle ABC must be right and so on.

[edit]
Correction: 5) needs more justification, see jollytall's comments below.
Moved the drawing file to this forum.
[/edit]

Title: Re: Triangle with angle split into 4 equal parts
Post by rloginunix on Mar 29^th, 2014, 5:41pm

The pattern that we see is that we keep extending each of the given objects. In "A Happy Medium" we extended the median. In "Bisector's Revenge" we extended the bisector. My gut is asking me now if we can extend the height and come up with "Height Expectations".

Any takers?

Title: Re: Triangle with angle split into 4 equal parts
Post by jollytall on Mar 30^th, 2014, 9:13am

rloginumix:
Thanks for the good drawings and elegant solutions. Nonetheless two little corrections:
First solution: Point 6 is obviously 180 and not 190 (typo)
Second solution: Point 5 is sort of incorrect. It is not necessarily true that angle D = beta. The "angle D" which is equal to beta is the MDB, but not necessarily ODB. It means that although ODB=OBD it does not mean that they also equal to beta. Obviously it is indeed true, but we have to use the fact that O is on DM line, simply because the centre of the circle must be on the perpendicular line of any spur's bisector, i.e. DM. From thereon, it is true, that ODB=MDB=beta.

The first solution can be a bit simpler. With the same logic you used to prove that BAC=BDC, we could rather say that ABD=ACD=beta. From the BHC triangle we know that HCB=90-beta. Hence BCD=90. The rest is the same.

Title: Re: Triangle with angle split into 4 equal parts
Post by rloginunix on Mar 30^th, 2014, 11:49am

Thank you for reviewing my work, jollytall. I've fixed the typo in solution 1 and added a comment to solution 2 that 5) needs more justification - I've got sloppy there even though the point that you duly noted is the very first proposition in Book 3.

Now that you've mentioned the solution 1 simplification I've spotted a similar one: instead of constructing DC we construct DA. AB then becomes a common segment for the angles ADB and ACB which are equal. From triangle BHC the angle ACB is 90 - beta, hence ADB = 90 - beta. But DBA = beta, hence, DAB = 90, etc.

Beautiful problem.

Title: Re: Triangle with angle split into 4 equal parts
Post by SWF on Mar 31^st, 2014, 6:10pm

Draw the triangle over the tessellation of regular octagons and squares. Points D, E and F respectively lie on the perpendicular, a bisector of the angle at A, and the bisector of segment BC. The angle at A is 90. The angle at B is 135/2=67.5. The angle at C must be 22.5.

Title: Re: Triangle with angle split into 4 equal parts
Post by towr on Mar 31^st, 2014, 11:23pm

That's just beautiful.

Title: Re: Triangle with angle split into 4 equal parts
Post by jollytall on Apr 1^st, 2014, 2:38am

SWF:
It really looks elegant and beautiful, but I am not sure I fully understand. If you make the drawing like this, then indeed A angle is 90, and all other conditions met. Does it prove that it is indeed 90?

I had proved relatively early that in case A angle is 90, it works perfectly (AM=MB on the earlier drawing, so beta = 90 - 3*beta, etc.). What I was not sure that it proves that this is the only solution. I was trying to move B and prove that then something goes wrong, but could not.
Now I feel the same: If you draw it on the octagons and squares like you did it, then it works and A angle =90. What is the extra step to prove that this is unique?

Thinking of a riddle where given a quadrilateral and the diagonals cross each other in right angle and half each other. What are the angles? I draw a square and prove it works, so the angles are 90. Obviously it does not prove that this is the only solution, any other rhomboid would work equally well.

Title: Re: Triangle with angle split into 4 equal parts
Post by rloginunix on Apr 1^st, 2014, 11:21am

SWF, you win my vote. Awesome idea.

I would like to add my modest (and newly found) solution, number four now. The gist of this one is to prove that the angle MBH = 2*beta = 45 degrees.

Solution 4, Bisector's Revenge: With a Vengeance.

http://www.ocf.berkeley.edu/~wwu/YaBBAttachments/rlu_4ea3.png

1). Triangle SBC is isosceles. Angles BHS and BHC are right by construction and definition of height. Angles HBS and HBC are equal by condition. The side BH is shared. From B1.P26 (ASA) triangles BHC and BHS are congruent. Hence, remaining angles and sides are equal correspondingly. What we will need from this is the fact that BC = BS and SH = HC.

2). Construct a circle Cir(B, BC) - centered at B with radius BC. It will intersect BA at point D. From the definition of circle it follows that BD = BC = BS.

3). By the definition of the median:

AM = MC = MS + SH + HC = MS + 2*SH

4). Apply B6.P3 (angle's bisector cuts the base proportionally) to triangle ABC:

AS/AB = SC/BC

AS = AM + MS = MC + MS = MS + 2*SH + MS = 2*(MS + SH)

SC = 2*SH

AS/AB = 2*(MS + SH)/AB = 2*SH/BC

(MS + SH)/SH = AB/BC

MS/SH + 1 = AB/BC

5). Apply B6.P3 (angle's bisector cuts the base proportionally) to triangle ABS since BM is its bisector by the given condition:

AM/AB = MS/BS, but from 2) we have: BS = BC, hence:

AM/AB = MS/BC

From 3) we have AM = MS + 2*SH, hence:

(MS + 2*SH)/AB = MS/BC or (rearranging AB and MS):

(MS + 2*SH)/MS = AB/BC

1 + 2*SH/MS = AB/BC, but in 4) we have AB/BC as MS/SH + 1, equate them:

1 + 2*SH/MS = MS/SH + 1, cancel out the ones:

2*SH/MS = MS/SH, substituting SH/MS = t:

2*t = 1/t or 2*t^2 = 1 or t = 1/sqrt(2) = SH/MS

But it is easy to prove that the angle MES is 90 degrees and SH/MS is the definition of the Cosine of the angle MSE, from which it follows:

angle MSE = arccos( sqrt(2)/2 ) = 45 degrees (ignoring cyclicity).

6). I am sure you, guys, can unwind it from here. The angle SME must be 45 degrees. ES = EM = ED. The angle BMD must be 45 degrees. DM must be parallel to BH.

7). The angles BMD and MBH are vertical and BMD = 45 = MBH = 2*beta or

2*beta = 45

and the rest of the solution follows.

7+1). An interesting property that we find is that Cir(B, BC) and Cir(M, MS) intersect at D.

[edit]
Moved the drawing file to this forum.
[/edit]

Title: Re: Triangle with angle split into 4 equal parts
Post by SWF on Apr 1^st, 2014, 6:38pm

jollytall, the tessellation does not prove that there is only one solution, but I interpreted the question as saying it is the case. I liked that approach better than the more straight forward trig solution:

Using the notation of rloginunix's last figure, except I will use "b" to represent angle beta. From sum of angles in a triangle=180:
Angle ACB=90-b, Angle BAC=90-3b, Angle AMB=90+2b

From Law of Sines (first for triangle ABC, then with triangle ABM):
AC/AB = sin(4b)/sin(90-b) = 2*sin(b)/sin(90+2b)
Factor of 2 in last numerator is from AM=AC/2.

Since cos(x)=sin(90-x)=sin(90+x), and sin(2x)=2*sin(x)*cos(x)...
sin(4b)*cos(2b)=2*sin(b)*cos(b)
2*sin(2b)*cos(2b)²=sin(2b)
cos(2b)²=1/2
The only solution with 0<4*b<180 is 2b=45 or b=22.5 degrees, but the sin(2b) term that was eliminated means b=0 would have been a solution if you count that as a triangle.

Title: Re: Triangle with angle split into 4 equal parts
Post by jollytall on Apr 1^st, 2014, 9:14pm

My original solution was also trigonometrical. I used tn(3b)-tn(2b)=tn(2b)+tn(b) from AM=MC or AH-MH=MH+HC.
From there I also got 22.5 (and 67.5, -22.5, -67.5). This was my full solution.

If we assume that there is only one solution, then without any complex logic, it is easy to prove that 22.5, 67.5 and 90.0 give a solution. If angle B=90, then M is the origo of the circle draws around the triangle, so AM=MB, i.e. BAM=ABM, 90-3b=b. b=22.5 fits for all the angles.

Title: Re: Triangle with angle split into 4 equal parts
Post by rloginunix on Apr 3^rd, 2014, 12:08pm

I think I've found an interesting little variation for those who like ruler and compass constructions, though of course it can be solved with trigonometry alone.

jollytall, do I have your permission to modify your problem statement slightly - one parameter only?

Title: Re: Triangle with angle split into 4 equal parts
Post by jollytall on Apr 3^rd, 2014, 9:06pm

Sure :-)

Title: Re: Triangle with angle split into 4 equal parts
Post by rloginunix on Apr 4^th, 2014, 8:52am

Thank you.

My rationale behind it was as follows. In the original problem the angles are fixed and they cut the base however they do. My thought was: let's reverse that. Let's fix the line segments on the base and let the angles be whatever they are:

Find the angles of (and construct) a triangle if it's same vertex height and bisector cut the half of it's base in three equal parts.

If you choose to construct the triangle assume that a given line segment MC is such that MC = MS + SH + SC and MS = SH = SC and the task is to reconstruct points A and B.

One construction that I've found locates A in [hide]two[/hide] steps and B in [hide]four[/hide] steps.

Title: Re: Triangle with angle split into 4 equal parts
Post by Annettagiles on Oct 23^rd, 2014, 4:51am

90,45,270 and 360

Title: Re: Triangle with angle split into 4 equal parts
Post by Grimbal on Oct 23^rd, 2014, 6:41am

on 10/23/14 at 04:51:26, Annettagiles wrote:

90,45,270 and 360

I dont't understand your answer. Could you please provide a picture?