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Topic: Another Triangle Construction (Read 2673 times) 

Barukh
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Another Triangle Construction
« on: Aug 5^{th}, 2014, 5:15pm » 
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In continuation to this thread... Construct triangle given: one side, the opposite angle, and the angle bisector to this side.


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dudiobugtron
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Re: Another Triangle Construction
« Reply #1 on: Aug 5^{th}, 2014, 6:39pm » 
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Some thoughts so far: The side and opposite angle (uniquely?) define a circle arc that the third vertex must be on. You can find the centre of that circle by creating a perpendicular bisector to the given side (which will be a chord of the circle, so therefore the centre must lie on that line). Then, since the angle at centre is twice the angle at the circumference, you can construct a line from that perpendicular 'centre line' at your given angle (assuming it is less than 90 degrees  otherwise use its supplement which is easy to construct) through either of the endpoints of your side, to find the circle's centre. Once you have found the centre of the circle, you know its radius, and so can construct the set of possible points that the third vertex can lie on. Now, you just need to use the length of the angle bisector (and the fact that is is a bisector) to figure out which of those points will work. I haven't worked out that part yet!


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rloginunix
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Re: Another Triangle Construction
« Reply #2 on: Aug 6^{th}, 2014, 7:07pm » 
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This is preparatory Part One (which may be used as a hint). It makes use of only the given side "a" and the angle "alpha" opposite to it. Since the side "a" is given it follows that we can locate triangle's two of three vertexes right away simply by cutting "a" on an arbitrary line. Let's name these vertexes A and B. So the objective then is to locate the remaining vertex C. Since the angle "alpha" opposite to "a" is given we can construct the locus of all the vertexes C from which the given line segment is visible under the given angle "alpha". I did it using B3.P20, central angle is double its circumference counterpart, but you can invent your own way. The construction steps are : 1). Line(Z, Y) perpendicular to AB : DA = DB (perpendicular bisector). 2). Line(A, X) perpendicular to AB. 3). Line(A, W) : angle XAW = alpha. 4). Line(A, W) intersects Line(Z, Y) at O = the circumcenter of the triangle ABC. Now, the vertex C must be somewhere on the arc AHB. To locate C exactly we need the second locus  the one that makes use of l(a)  the given bisector. In the future drawing(s) I will skip the intermediate points of this drawing and will only show the relevant ones.


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dudiobugtron
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Re: Another Triangle Construction
« Reply #3 on: Aug 6^{th}, 2014, 7:36pm » 
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rloginunix's method is cool because it works even if alpha is equal to or greater than 90.


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rloginunix
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Re: Another Triangle Construction
« Reply #4 on: Aug 7^{th}, 2014, 3:09pm » 
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This is Part Two, main idea. It's a little messy but bear with me. I'll take it one step a time. I've highlighted all the points that we can construct with the measures given in one color and the point sought after, C, in another. For the second locus our only hope is to use the given l(a). From the preparatory drawing we see that it would be ideal to somehow construct the line segment equivalent in length to GC. Clock pendulums usually swing at the "bottom" so for this experiment imagine a reverse pendulum GC swinging about the point G with two circles on it: Cir(G, GF) and Cir(E, EC = EF = l(a)/2) (clockwise). When the same pendulum swings counterclockwise you have the Cir(M, MN = l(a)/2). The interesting point about this is that the center of the given bisector, E, will eventually intersect HB at J and HA at M. Since everything that follows hinges on that observation I'd imagine some high brow math must be used to prove that. However, at the moment I'm not sure how to go about this proof.. Another way to think about this is to imagine a Cir(E, EC) rolling without slipping along the fixed Cir(G, GF). Either way it suggests a construction.


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rloginunix
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Re: Another Triangle Construction
« Reply #5 on: Aug 8^{th}, 2014, 8:20am » 
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This is Part Three, the construction. So we pick up where we left off at Part One. All the steps for Part One are carried out, then: 1). Bisect the given l(a). 2). Line(B, H). 3). Cir(B, l(a)/2). 4). Line(B, H) and Cir(B, l(a)/2) intersect at J. 5). Line(G, J). 6). Cir(J, JB). 7). Line(G, J) and Cir(J, JB) intersect at K. 8). Cir(G, K). 9). Cir(G, K) intersects the circumcircle at C and C'. So it looks like this method yields two similar triangles. Of course you can do the same if you use the line segment AH to build up the equivalent of GC. I shall be posting the proof that this construction works.


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rloginunix
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Re: Another Triangle Construction
« Reply #6 on: Aug 8^{th}, 2014, 12:51pm » 
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This is Part Four, the proof. I've regiggered the drawing a bit to make it more evident where does the point P come from  the intersection of Line(G, J) and Cir(J, JB). The task is to prove that GK = GC. 1). Based on AAA the triangles FDG and GCH are similar: GF/GD = (2R)/(GF + l(a)) or GF(GF + l(a)) = 2R*GD. 2). Based on B3.P36 from the Cir(J, JB) and a tangent GB we have: GP*GK = GB^2 or GP(GP + l(a)) = GB^2. 3). However, based on AAA the triangles GDB and GBH are similar: GB/GD = (2R)/GB or GB^2 = 2R*GD = GP(GP + l(a)) (from 2).). 4). From 1) and 3) we have GF(GF + l(a)) = 2R*GD = GP(GP + l(a)) or GF(GF + l(a)) = GP(GP + l(a)) and hence GP = GF. But FC = l(a) = PK by construction, hence GK = GC.


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Barukh
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Re: Another Triangle Construction
« Reply #7 on: Aug 9^{th}, 2014, 7:08am » 
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Nice analysis, rloginunix! I hope you liked the problem. However, there is one thing I am probably missing: on Aug 8^{th}, 2014, 12:51pm, rloginunix wrote:But FC = l(a) = PK by construction, . 
 The second part of the equation is clear, but I don't see the first part in your construction.


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rloginunix
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Re: Another Triangle Construction
« Reply #8 on: Aug 9^{th}, 2014, 9:53am » 
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on Aug 9^{th}, 2014, 7:08am, Barukh wrote:I hope you liked the problem. 
 Yes, I did. Thank you for creating an entire series of interesting problems. on Aug 9^{th}, 2014, 7:08am, Barukh wrote:but I don't see the first part in your construction. 
 Will the proof work: By construction HG is a perpendicular bisector of AB. Hence by SAS the triangles GDA and GDB are congruent => GA = GB => the circumference angles GCA and GCB subtend equal segments => by B3.P21 (In a circle the angles in the same segment equal one another) angle GCA = angle GCB => FC must indeed be a given bisector?


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