Limit as a line segment's length approaches 0.

[Matthew Ko]

Hi,
This question does have components of geometry but it's from my Calculus class, so I'm initially posting it here. I'm literally stuck with this question from the beginning and the textbook apparently has no answer keys for this particular question. The question and my attempted work are attached. Thank you for your time in reading this post and helping me.

 

[Deleted member 4993]

Hi,
This question does have components of geometry but it's from my Calculus class, so I'm initially posting it here. I'm literally stuck with this question from the beginning and the textbook apparently has no answer keys for this particular question. The question and my attempted work are attached. Thank you for your time in reading this post and helping me.

If I were to do this problem, I would do part 'b' first. I would assign co-ordinates to B (0,0) and to C (0,d) and assume mABC = \(\displaystyle 2\theta\)

Now I would calculate the co-ordinate of A and P as functions of \(\displaystyle \theta\)

and continue.....

 

[HallsofIvy]

I would say that, as AM decreases, everything drops straight down. Since the two angles at B are equal, P is the midpoint of AC and winds up on BC at the midpoint of MC.

 

[Matthew Ko]

Thank you so much for your help.

I figured that (1) if I draw a line from P down to MC (I called that point N) that is perpendicular to that line and the line PM, AND (2) show that △PMC∽ △ABC (latter being an isoleses triangle), I can call △PMC also an isoleses triangle and, therefore, can prove that P indeed winds down to the midpoint of the line MC (I need this extended proof, by the way, because my assignment is to delineated it in my project report).

I could prove that △PNC∽ △AMC because (1) ∠ ACM=∠ PCN and (2) ∠ CAM=∠ CPN since PN || AM (AA similarity). I couldn't, however, prove that ∠CAB=∠CPM as I couldn't prove PM || AB (which would allow me to conclude that △PMC∽ △ABC). Nor could I prove that ∠CPN=∠MPN, which would allow me to say △PNC≡ △AMC (ASA congruence). My work is attached below. May I receive further help from this point? Thank you so much for your time and consideration.

 

[Dr.Peterson]

I would say that, as AM decreases, everything drops straight down. Since the two angles at B are equal, P is the midpoint of AC and winds up on BC at the midpoint of MC.

No, P is not the midpoint of AC. The locus of P is a curve.

@Matthew Ko: You may recall from geometry a theorem about the location of P on an angle bisector (that is, the ratio AP : AC); if not, you may be able to prove it based on the areas of the triangles above and below AP. That leads to a nice answer to the question.

 

[Matthew Ko]

Dear Dr. Peterson,

Thank you for your help. However, I am unsure what the bisector angle theorem would lead me to. Will it allow me to prove the similarity I was trying to prove or something different that would help me answer the question? Do you entirely discard the notion that point N (on my work) is the midpoint of CM? I am also unsure about the expression "above and below AP." Perhaps that means another triangle that I should draw outside the triangle ABC?

 

[Dr.Peterson]

Dear Dr. Peterson,

Thank you for your help. However, I am unsure what the bisector angle theorem would lead me to. Will it allow me to prove the similarity I was trying to prove or something different that would help me answer the question? Do you entirely discard the notion that point N (on my work) is the midpoint of CM? I am also unsure about the expression "above and below AP." Perhaps that means another triangle that I should draw outside the triangle ABC?

Here is the theorem I had in mind (with a different proof than the one I hinted at): https://mathbitsnotebook.com/Geometry/Similarity/SMAngle.html

It is also stated here, with a generalization that makes the proof more complicated than it has to be (no trig is needed): https://en.wikipedia.org/wiki/Angle_bisector_theorem

The triangles I referred to are ABP and CBP; I think I was rushing too much to go back to the image and check. My proof is different from that on the first page I just referred to, but similar to proof 3 in Wikipedia.

N is definitely not the midpoint of CM, because the ratio AP : PC is not 1 : 1.

 

[Matthew Ko]

Thank you so much for your help. I appreciate your time.

I did know initially that you were referring to the angle bisector theorem. However, I wasn't quite sure how the ratios I get from the theorem would help me answer my question. It doesn't help me to define PB as dependent on one variable (I'm trying to make all the coordinates in the graph totally dependent on constant values like d [or CB], not variable values, as Subhotosh advised above) nor any other lines like AC or CP.

 

[Matthew Ko]

I figured out with some help that AP=AC-CP and set a limit of CP as AC approaches (BC/2), so (a.) is quite settled. Thank you so much for your help.

However, as an additional note, I am not quite sure how to solve (b.). I defined the coordinates of A and P (seen in the picture attached in the post above), but even if they're right, how can one form an equation out of coordinate equations (meaning, coordinate values defined by variables as P in my work)? And if the locus of P is a curve, how can I know its degree and prove it as well?

 

[Dr.Peterson]

The goal will probably be to express the coordinates of P in terms of a parameter (I'd use MA = t, rather than your theta, but it ought to work either way), and then eliminate the parameter to get an equation relating x and y.

It doesn't look like you've used the fact that P is on AC.

You haven't stated your answer to (a), but I'll hope you got it right.

 

[Dr.Peterson]

I was able to obtain a fourth-degree equation that matches my GeoGebra construction, so I think it's correct. I took P(x,y) and A(0,t) as given and wrote equations expressing AP^2/AB^2 = AC^2/BC^2 and P lies on AC, then eliminated t.

But it isn't exactly something I'd want to "sketch". I used Desmos. It looks a lot like a hyperbola.

 

[Dr.Peterson]

Just to let you know, it turns out that it is a hyperbola, with its vertices and one focus at known places, though I don't think one could guess any of this. There must be an easier way to find the equation than what I did, as it does not end up really being fourth-degree.

 

[Matthew Ko]

Thank you so much for all your time and attention, sir! All your help was of great avail in working with my project.