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Old 2007-02-04, 21:11   #1
philmoore
 
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Default Angle bisector problem

Consider a right triangle ABC with the right angle at point C. Let AD, with D on BC, be the angle bisector of the angle at A (i.e., angle BAC) and let BE, with E on AC, similarly be the angle bisector of the angle at B, angle ABC. Given that AD is of length 9 units, and BE is of length 8\sqrt2 units, the problem is to find the length of the hypotenuse AB.
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Old 2007-02-07, 08:20   #2
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Label AB=c, AC=b, BC=a, CD=x and CE=y.
Applying pythagore thm to triangles ABC, EBC and ADC give the following three equations:

a^2+b^2=c^2
a^2+y^2=128
b^2+x^2=81

Using the bisector theorem we get the following two equations:

x / b = (a-x)/c
y / a = (b-y)/c

Rewriting the latter two equations we get

x=ab/(b+c)
y=ab/(a+c)

Plugging these results in the first three equations we get

a^2+b^2=c^2
a^2+a^2 b^2/(a+c)^2=128
b^2+a^2 b^2/(b+c)^2=81

The second equation rewrites to

b^2=(128-a^2)(a+c)^2 / a^2

Plugging into equation 1 and 3 gives some mess I have not cleaned out yet, but all in all it gives two equations in the unknowns a and c, eliminating a from one of the equations finally gives an equation for c, don't know whether this approach works for sure, but it is a decent try


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Old 2007-02-07, 12:18   #3
mfgoode
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Cool From a colleague.


I gave an associate of mine, who teaches maths and is a math wizard, and he said that this once again turns out to be an irredicuble cubic and suspects the dimension 8 rt.2 is not correct which makes the solution very difficult if not impossible. I took his word for it and so well left it alone without even a trial.

He may be wrong but I would like to see a solution by the poster himself if no one else can give it. Maybe the trig method might work.

Mally

Last fiddled with by mfgoode on 2007-02-07 at 12:19
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Old 2007-02-07, 18:45   #4
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The approach of Kees definitely can be used to solve the system graphically, as that is what I did when I first encountered this problem at a math teachers conference last April. Because multiple choice answers were offered, I was able to eliminate the wrong choices, but was intrigued by the simple form of the correct answer, and did not actually find a correct algebraic solution until recently. I suspect that Kees' equations can lead to an algebraic solution, but I still don't see how.

As for the opinion of Mally's associate, it is wrong, but does have a grain of truth to it. If we replace 9 and 8\sqrt2 by other values, we often do arrive at an irreducible polynomial, but these particular values make the problem more easily solvable.

If anyone would like a hint, I will outline a method and let you have the fun of working out the details:

Note that the two angle bisectors must cross at an angle of 135 degrees. I obtained a solution through the use of coordinates, placing the vertex A at the origin and the point D on the x-axis at (9,0).

And in case you get stuck, another hint:

The point C is on the line y = mx and the point B is on the line y = -mx. You will obtain a cubic equation in m with one real root and two complex roots. For the particular lengths of the angle bisectors in the given problem, the real root is rational.
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Old 2007-02-08, 07:45   #5
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Quote:
Originally Posted by philmoore View Post
The approach of Kees definitely can be used to solve the system graphically, as that is what I did when I first encountered this problem at a math teachers conference last April. Because multiple choice answers were offered, I was able to eliminate the wrong choices, but was intrigued by the simple form of the correct answer, and did not actually find a correct algebraic solution until recently. I suspect that Kees' equations can lead to an algebraic solution, but I still don't see how.

As for the opinion of Mally's associate, it is wrong, but does have a grain of truth to it. If we replace 9 and 8\sqrt2 by other values, we often do arrive at an irreducible polynomial, but these particular values make the problem more easily solvable.

If anyone would like a hint, I will outline a method and let you have the fun of working out the details:

Note that the two angle bisectors must cross at an angle of 135 degrees. I obtained a solution through the use of coordinates, placing the vertex A at the origin and the point D on the x-axis at (9,0).

And in case you get stuck, another hint:

The point C is on the line y = mx and the point B is on the line y = -mx. You will obtain a cubic equation in m with one real root and two complex roots. For the particular lengths of the angle bisectors in the given problem, the real root is rational.


Quote:
Originally Posted by philmoore
Because multiple choice answers were offered, I was able to eliminate the wrong choices, but was intrigued by the simple form of the correct answer, and did not actually...
Well philmoore what is the 'simple form of the correct answer' ?
Please give us your cubic eqn. so we can tackle it.

Your hints are very elementary and the fact you have resorted to a graphical solution and then by co-ordinate geometry and finally an algebraic one deters me from even tackling it!

And why do you choose Kees solution over Davieddys? Davy's in my opinion is
the one that can give all 3 real values.

I have on hand all possible solutions for the roots but they are laborious to write down and I dont use TEX.

So philmoore please dont meander your way thru a seemingly simple geom problem. Lets have some concrete hard-boiled roots to this problem as you are the one who started it and so you should end it.

Mally

Last fiddled with by mfgoode on 2007-02-08 at 07:54
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Old 2007-02-08, 08:46   #6
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working out the final details of my proposed solution, after eliminating x and y we set

a= c cos t
b= c sin t

we can then easily eliminate c to get an equation in (cos t, sin t)
Setting now

sin t=p
cos t=sqrt(1-p^2)

we can arrive at an polynomial equation in p.
Not having enough confidence (well, I do have enough confidence, but I would like to give the right answer straight away) in handling this kind of operations myself, I will work it out in PARI and give the equation in a next posting
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Old 2007-02-08, 08:52   #7
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Quote:
Originally Posted by mfgoode View Post
And why do you choose Kees solution over Davieddys?
Well, and this is just a wild stab in the dark, maybe he chose Kees's solution over Davieddy's because Kees has actually posted one whereas Davieddy has yet to contribute to this thread. Maybe you just made a typo although, given that in other threads you have equated typos with deliberate attempts to mislead people, I doubt you'll make such a claim.
I appreciate that other people may find this response a tad harsh, but I haven't been on here very long and I have already had enough of this pompous, condescending windbag.
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Old 2007-02-08, 09:14   #8
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PARI came up with a numerical solution, so I do not have it in radical form.

After finding a equation in p (of degree 8 (!)) we finally find

AB=13.16

Not very satisfied with this solution, hoped that the substition would yield an easier equation, but none of that happened.
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Old 2007-02-08, 11:10   #9
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Degree 8 is definitely an improvement on my degree 12 equation in c, although that equation was actually degree 6 in c-squared!

Try this, following up on the earlier hints:

Drop a perpendicular from the point D at (9,0) to the line y = - mx and find the intersection of these two lines to get the point B, in terms of m. The intersection with y = mx will give us point C. The angle bisector of ACB, as noted above, intersects the x-axis at a 45 degree angle and hence has a slope of 1. Find the intersection of this line with y = - mx now to locate point E. We now know A and E in terms of m, and equating the distance between these points to 8.sqrt(2) gives us a cubic equation in m. This equation factors, and has one rational root, m = 1/2, and two complex roots. From the rational root, we easily find the hypotenuse AC.

Kees, I haven't checked details, but I suspect your method will work better if you use p = tan t, instead of p = sin t as your variable. But notice that I am using m = tan (t/2) as the variable in my solution.
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Old 2007-02-08, 12:19   #10
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I think that you are right,

using the substition p = tan t

and therefore cos t= 1/(p^2+1), sin t=p^2/(p^2+1)

I stumble also upon an equation of degree 12 (in p^2) and therefore an equation in degree 6. But I do not see a solution to that equation.
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Old 2007-02-13, 17:38   #11
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Cool Windbag

Quote:
Originally Posted by DJones View Post
Well, and this is just a wild stab in the dark, maybe he chose Kees's solution over Davieddy's because Kees has actually posted one whereas Davieddy has yet to contribute to this thread. Maybe you just made a typo although, given that in other threads you have equated typos with deliberate attempts to mislead people, I doubt you'll make such a claim.
I appreciate that other people may find this response a tad harsh, but I haven't been on here very long and I have already had enough of this pompous, condescending windbag.


Well Djones you blabber mouth learn to keep your trap shut before you make any personal comments. Its a sure sign of an inferiority complex.

You are just like the second, second in a heavy weight bout.- you know the one who replaces the teeth guard in a champions mouth at the start of the round.?

Why dont you put on a Scottish tweed checked mini skirt and become a cheer leader ? It will suit you better I tell you I will be the first to go for you!

FYI: the two puzzles on Quadrilaterals and this one are closely connected if you have any sense in your head to have followed both before you came into the ring.

For your own benefit kindly read posts no.s 26 and 28 where Davy makes his entrance in the Quadrilateral problem and gives the method to solve for real values. Yes read all about it !

Wind bag ? I feel like punching you as a sand bag I used in practice boxing.

But I will save my verbal knuckles for another encounter with you, I can assure you!

Mally
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