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Malakor

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Aug 12, 2018
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I've been stuck on this problem for a while now, so I thought I'd try to get help online. I'd be incredibly thankful for any pointers you could give me to get started, since I've tried approaching it in several ways but been unable to get an answer.

"The point P is on the bisector of an angle with its corner in A. A line is drawn through P that intersects the angles legs in points B and C. Show that the value of 1/AB + 1/AC is the same, regardless of how we choose to draw the line."
 
I've been stuck on this problem for a while now, so I thought I'd try to get help online. I'd be incredibly thankful for any pointers you could give me to get started, since I've tried approaching it in several ways but been unable to get an answer.

"The point P is on the bisector of an angle with its corner in A. A line is drawn through P that intersects the angles legs in points B and C. Show that the value of 1/AB + 1/AC is the same, regardless of how we choose to draw the line."
Are you familiar with laws of sines and cosines for a triangle?
 
Are you familiar with laws of sines and cosines for a triangle?

To some extent, yes, although I'm having trouble applying either in this particular case. I thought I had it after drawing two lines (from B and C) at a right angle to the bisector and using the law of sines to prove the statement but I'm having trouble reaching the conclusion that the number 1/AB + 1/AC is independent of the leaning of the line BC...
 
… an angle with its corner in A …
I read "corner" to mean vertex. Symbol A represents a point.


… [a line] intersects the angles legs in points B and C.
Symbols B and C also represent points.

Also, an angle has rays, not legs.

A right triangle has legs (the two shorter sides). Are you working with a triangle?


Show that the value of 1/AB + 1/AC is the same, regardless of how we choose to draw the line.
I don't understand this notation. Symbols A,B,C represent points. Are you using the same symbols to also represent lengths of line segments?

Standard notation uses upper-case letters for angles (vertices) and lower-case letters for lengths. In this diagram, we see the correspondence between angle labels and length labels. That is, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C.

triABC.JPG


:idea: Posting a quick sketch of your situation would be very helpful.

PS: You don't need to submit duplicate posts. The forum guidelines explain why, in this notice.
 
I apologize for my lack of clarity (and the double post, thought it didn't go through the first time), English is not my first language and I did my best translating my problem. I did not know that there was a difference between vertex and point, but after searching around, I probably mean vertices in this case. I definitely meant Rays!

I'll rephrase the problem:

"The vertex P is located at some point on the bisector of an angle with its corner in A. A line is drawn through P that intersects the angles rays in points B and C. Show that the value of 1/c + 1/b is the same, regardless of how we choose to draw the line (after choosing the vertex P)."

GeometryProblem.jpg
 
… I did not know that there was a difference between vertex and point …
No, they are the same. A vertex is a point.

I tried to say that a point is not the same as a line segment. I wrote that because I had read your AB as \(\displaystyle A \cdot B\) instead of \(\displaystyle \overline{AB}\). My apologies, for being dense. (I have a small hangover.)

Can you answer Subhotosh's question? If not, can you explain one or two of your approaches? If nobody has replied after dinner, I'll take another look. :cool:
 
No, they are the same. A vertex is a point.

I tried to say that a point is not the same as a line segment. I wrote that because I had read your AB as \(\displaystyle A \cdot B\) instead of \(\displaystyle \overline{AB}\). My apologies, for being dense. (I have a small hangover.)

Can you answer Subhotosh's question? If not, can you explain one or two of your approaches? If nobody has replied after dinner, I'll take another look. :cool:

I already answered him further up, or was it not clear enough?

Anyway, the methods that I've tried and which are explained in the course litterature have been a combination of Thales's Theorem and the Angle Bisector Theorem as well as the fact that two triangles who share a base have an area relation equal to their height relation; \(\displaystyle area_a / area_b = height_a / height_b\). I've tried drawing extra lines (see image) to get the required equivalencies but can't seem to reach the conclusion that \(\displaystyle 1/c + 1/b = 1/c_2 + 1/b_2 \) for two randomly chosen rotations of the line. I know that \(\displaystyle \triangle ABD \sim \triangle ACE\) and \(\displaystyle \triangle BDP \sim \triangle CEP\) and therefore \(\displaystyle \overline{AB} / \overline{AC} = \overline{BD} / \overline{CE} = \overline{AD} / \overline{AE}\) and \(\displaystyle \overline{BD} / \overline{CE} = \overline{DP} / \overline{EP} = \overline{BP} / \overline{CP}\) and I've tried using replacement and algebraic calculations to reach a conclusion that \(\displaystyle 1/c + 1/b\) is solely dependent on \(\displaystyle \overline{AP}\) but with little luck.

EDIT: Forgot to upload the new image:

GeometryProblem.jpg
 
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… I already answered him further up, or was it not clear enough? …
Oh, I thought he had asked you to show your paper work (his usual boilerplate response).

I see that you're heavy into classical geometry (not my forte). But, you've shown enough now, for others to understand what's going on, so I'll bow out.
 
I would set it up this way

I tested the conjecture on GeoGebra and it seems to be true.
Without loss of generality, I fixed ...
P at (1,0) and A at (0,0)
I defined point H to represent the spread of the rays.

This yields nice equations for the 3 rays.

AB: y = hx
AP: y = 0
AC: y =-hx

Capture.JPG

Your job is to find C by intersection of rays BP and AC.
Find AB and AC in terms of h and B.
Find AH in terms of h.
Show 1/AB + 1/AC = 2/AH
 
Simpler Approach

It seems like there should be a simpler or even trivial approach.
I looked way too long for an approach to use the two ladder problem
and/or the construction of the HM by Pythagoras.

It seems this brute force method (above) is fastest for aged math
practitioners like myself.
 
Last edited by a moderator:
It seems like there should be a simpler or even trivial approach …
I thought that, also. Maybe it's simple, using Thales Theorem (and, perhaps, other things mentioned in post #7). I'm ignorant of most classical geometry; the doctor may have some suggestions.
 
Anyway, the methods that I've tried and which are explained in the course litterature have been a combination of Thales's Theorem and the Angle Bisector Theorem as well as the fact that two triangles who share a base have an area relation equal to their height relation; \(\displaystyle area_a / area_b = height_a / height_b\). I've tried drawing extra lines (see image) to get the required equivalencies but can't seem to reach the conclusion that \(\displaystyle 1/c + 1/b = 1/c_2 + 1/b_2 \) for two randomly chosen rotations of the line. I know that \(\displaystyle \triangle ABD \sim \triangle ACE\) and \(\displaystyle \triangle BDP \sim \triangle CEP\) and therefore \(\displaystyle \overline{AB} / \overline{AC} = \overline{BD} / \overline{CE} = \overline{AD} / \overline{AE}\) and \(\displaystyle \overline{BD} / \overline{CE} = \overline{DP} / \overline{EP} = \overline{BP} / \overline{CP}\) and I've tried using replacement and algebraic calculations to reach a conclusion that \(\displaystyle 1/c + 1/b\) is solely dependent on \(\displaystyle \overline{AP}\) but with little luck.

I tested the conjecture on GeoGebra and it seems to be true.
Without loss of generality, I fixed ...
P at (1,0) and A at (0,0)
I defined point H to represent the spread of the rays.

This yields nice equations for the 3 rays.

AB: y = hx
AP: y = 0
AC: y =-hx

View attachment 9931

Your job is to find C by intersection of rays BP and AC.
Find AB and AC in terms of h and B.
Find AH in terms of h.
Show 1/AB + 1/AC = 2/AH

Let's combine these ideas. Add to your picture the segment XY through P perpendicular to AP, to make an isosceles triangle AXY. You want to show that 1/AB + 1/AC = 1/AX + 1/AY.

For simplicity, I labeled segments AC=b, AB=c, BX=x, CY=y, AX=AY=d, PB=e, PC=f.

Apply the angle bisector theorem to ABC, as you have done; AC:AB = PC:pB, i.e. b/c = f/e.

You want to show that 1/b + 1/c = 2/d, or equivalently cd + bd = 2bc. Gathering terms with b on the left and with c on the right, bd - bc = bc - bd. See if you can turn this into a proportion that suggests a pair of similar triangles. To actually see that pair, you'll have to construct another segment.

I'll leave it at that so you can have the thrill of discovery.
 
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