Triangle area/construction by angle bisectors (open pr.).

[ins-]

1. Let l_{a}, l_{b}, l_{c} are angle bisectors of triangle ABC. Express the area of ABC by them.
2. I'm wondering also how the triangle may be constructed by three angle bisectors.

[Юстас]

Well, the formula for the length of the bisector as a function of the sides is known, you can easily get it yourself. Then, we have a system of 3 equations..But may be it does not have any good solution in terms of elementary functions.

[ins-]

Yes, I think so. I wrote this system ... but nothing good. I'm surprised because I know how to express triangles surface/area by its medians, heights/altitudes but ... angle bisectors... I think it is very interesting problem and if I have good luck someday I may solve it. I'm thinking that it we may not need to solve this system. See the following problem:
$ a, b, c, d - reals $
$ a^{2}+b^{2}=1 $
$ c^{2}+d^{2}=1 $
$ ac + bd = 0 $
Find: $ ad-bc $
They gave me when I was 8-th grade on olympiad it is not need to know exact values of a,b,c,d to calculate the expression. But you need to be very good in algebra. I think we may select some form for the surface formulae and then using not defined coefficients or something similar we may calculate the surface. But we need to be very very good with algebra...and with hypothesis definitions. I think so some geometric approach may be very effective (sometimes with geometric approaches we get facts that are almost impossible to get with algebra)... maybe additional constructions but I have no idea about it.
What about constructive problems?

[незваный гость]

ins- писал(а):

I'm wondering also how the triangle may be constructed by three angle bisectors.

The triangle is not defined by bisectors.

[ins-]

Are you sure? I heard somewhere that triangle is defined by any 3 linear elements. Am I wrong?

[незваный гость]

ins- писал(а):

But you need to be very good in algebra.

It is obvious: Let $a = \cos\varphi$, $b = \sin\varphi$, $c = \cos\psi$, $d = \sin\psi$. Then the problem reduces to $\cos(\psi-\varphi) = 0$, find $\sin(\psi-\varphi)$. Indeed, it is either $1$ or $-1$.

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ins- писал(а):

Are you sure? ‹…› Am I wrong?

yes to both questions

[ins-]

There is Lagrange equality it may be used did you read - it is 8 grade problem?
sqrt(1)*2)-3)*3)) may be used (I'm sorry I don't use tex). I also discovered your solution but it is more limited. It doesn't works with problems like:
a, b, c, x, y, z - reals
a^{2} + b^{2} + c^{2} = 25
x^{2} + y^{2} + z^{2} = 36
ax + by + cz =30
Find value of:
(a + b + c)/(x + y + z)

How you can prove that the triangle is not defined by its angle bisectors?

[незваный гость]

ins- писал(а):

it is 8 grade problem?

I did. It does not mean one have to be ignorant in 8th grade.

ins- писал(а):

sqrt(1)*2)-3)*3)) may be used (I'm sorry I don't use tex)

It is up to you (more or less). It is illegible so I have no choice but to ignore.

ins- писал(а):

I also discovered your solution but it is more limited.

All solutions are limited. Esp. beautiful ones… Why to bother…

Добавлено спустя 2 минуты 55 секунд:

ins- писал(а):

How you can prove that the triangle is not defined by its angle bisectors?

Folklore?!

Добавлено спустя 7 минут 45 секунд:

ins- писал(а):

It doesn't works with problems like:

Consider vectors $A=(a,b,c)$, $X=(x,y,z)$. $|A|=5$,$|X|=6$, $(A,X)=30=|A||X|$. Therefore, $A$ and $X$ are collinear and the ratio is equal to $|A|/|X|$.

It is interesting. Apparently you love geometry but you do not see it in two of your recent problems.

[ins-]

At the beginning of 8-th grade people in Bulgaria (this round of the olympiad was in this period) know only up to middle line in the triangle material. Your solution is possible for people in the beginning of 10 grade. No one not say it is not original or beuautiful. But it works only in case when in first three equations we have in right side 1's not 2, 3, 5 ... etc. - in this case your substitution is impossible. 1), 2), 3) and sqrt in: sqrt(1)*2)-3)*3)) are equality 1), 2), 3) respectively I tried do say
(a^{2}+b^{2})(c^{2}+d^{2})=(ac+bd)^{2}+(ad-bc)^{2}. It is known as Lagrange equality. Your answer is correct.
The second problem may be solved by using 1)*2)-3)*3). You gave again a good solution. (answer is 5/6).
Goal here is not to solve this problems I wanted to show that sometimes when something is 'invisible' - it may be found. I'm not sure for the area of the triangle by three angle bisectors but I think it may be found. I asked you to prove that the triangle is not defined by three angle bisectors because it is defined by three sides, three medians, three heights (three linear elements) why not to be defined by three angle bisectors? May you or someone else prove your statement? I think if it is true it will be very interesting.

Did you try this problem: http://dxdy.ru/viewtopic.php?t=9676?

[незваный гость]

ins- писал(а):

1), 2), 3) and sqrt in: sqrt(1)*2)-3)*3)) are equality 1), 2), 3) respectively

You are extremely innovative. This is first time in since my first day in school (September 1, 19xx) that somebody used 1), 2), 3) to designate objects in a mathematical expression. Most people use letters for that purpose, with a fringe benefit of parentheses matching. Of course, your way It is much easier to understand than cryptic $\sqrt{p q - r^2}$.

ins- писал(а):

I wanted to show that sometimes when something is 'invisible' - it may be found.

And may be not. And the probability is that most problems are ugly (very much like most numbers are transcendental). So, unless you can demonstrate nice solution, please, please do not brag that it exists. Or that you believe that it exists.

ins- писал(а):

Did you try this problem…

No. It is boring. I see no value in it.

[ins-]

'transcedental' is very interesting word. At the moment no one of us cannot solve this problem, but it doesn't mean it have no solution or it have solution. As it is written it is an open problem. It come to show - even with strong traditions and thousand years of history in math there are too much not solved problems. Intention of this problem is to discover something not discovered by anyone. In my opinion a triangle is defined by any 3 elements. Sometimes it is hard to believe. For example - try to solve the following problem:

New problem:
a) Constuct triangle ABC by given angles: CAB, CBA and median from C to AB (let say - CM - M is from AB such that AM=MB).
b) Find the Surface/Area of the triangle by these elements.

I can solve this problem I just composed it. And I'm not wrong for the solution of my new problem. If you cannot solve it I'll post a solution. I know the triangle is defined by any 3 elements from my school teacher in math. If I have a good luck and my problem is solvable I'll solve it.

About my new problem for equaliteral triangle - it is a hard and not standard problem. It is so boring as some of IMO problems. It is just my opinion. My goal is not confrontation but collaboration. I'm posting open problems because I want people to be more creative and try to think for problems outside of their textbooks and homeworks.

[незваный гость]

Tried. Solved.
1) Lets take an arbitrary $c^*$ and construct a triangle from it and two given angles. Obviously it is similar to the one we need. So we can measure its median $m^*$ and construct $c$ from proportion $c:c^*=m:m^*$.

BTW, the approach works with any “two angles, one linear element” problem.

2) I do not see anything but exercise in typing here. The area can be expressed thru $R^$ and angular functions, the square of median — too.

ins- писал(а):

I know the triangle is defined by any 3 elements from my school teacher in math.

Oh, those authorities. Try a triangle $a$, $m_a$, $h_a$. Or a triangle $a$, $h_a$, $\beta_a$.

[ins-]

незваный гость писал(а):

:evil:
Tried. Solved.
1) Lets take an arbitrary $c^*$ and construct a triangle from it and two given angles. Obviously it is similar to the one we need. So we can measure its median $m^*$ and construct $c$ from proportion $c:c^*=m:m^*$.

BTW, the approach works with any “two angles, one linear element” problem.

2) I do not see anything but exercise in typing here. The area can be expressed thru $R^$ and angular functions, the square of median — too.

ins- писал(а):

I know the triangle is defined by any 3 elements from my school teacher in math.

Oh, those authorities. Try a triangle $a$, $m_a$, $h_a$. Or a triangle $a$, $h_a$, $\beta_a$.

You are right my problem is not difficult but it is not 'obvious' that the triangle is defined by these elements. a ) is from a book. b ) is composed by me I don't took in mind this I used that $cotg\phi = 0.5(cotgA-cotgB) $ Where $ \phi $ is the angle between $ m_c $ and AB. If you want something difficult - solve my open problem 1 - 'beautiful and difficult'.

First triangle is defined by its 3 elements algorithm for construction:
1. Construct a.
2. Draw a line g parallel to a and distance between a and g is $ h_a $.
3. Draw a circle k with center middle of A and with radius $ m_a $.
4. A is intersection point of k and l.
In this case we may construct 0, 1, 2, 4 triangles but no matter how they are if it exists it is the same triangle.
I don't know what you want to say with $ \beta_a $ for your second example.

[незваный гость]

ins- писал(а):

First triangle is defined by its 3 elements algorithm for construction:

Yep. I guffed here. For whatever reason I thought one can slide the side along the line perpendicular to the height.

ins- писал(а):

I don't know what you want to say with $ \beta_a $ for your second example.

$\beta$ is a more or less traditional letter for an angle bisector in Russia. It does not matter, although: I guess, it is the same guff.

ins- писал(а):

You are right my problem is not difficult but it is not 'obvious' that the triangle is defined by these elements.

Obviousness depends on solver.

ins- писал(а):

is from a book.

I have never seen any problem of this kind in textbooks. So what? There is exactly 48 problems involving sides, heights, medians and bisectors. Will we “invent” all of them?!

[ins-]

Цитата:

ins- писал(а):

You are right my problem is not difficult but it is not 'obvious' that the triangle is defined by these elements.

Obviousness depends on solver.

ins- писал(а):

is from a book.

I have never seen any problem of this kind in textbooks. So what? There is exactly 48 problems involving sides, heights, medians and bisectors. Will we “invent” all of them?!

1. Solver is intended to be a pupil at level more than the average.

2. We will not need to invent all such problems only the interesting. As I said I reinvented the wheel 5-6 times. But it is not made only from me. I took a look at the Bulgarian Math Olympiad 2003 - III round - 1-st problem. This problem is present in a standard Bulgarian book for student preparation exam even it is more general in the textbook. Pretention of these people are that all problems in BMO are new, and not known. I think also no one know all problem ever discovered. But why not if a problem is new and interesting and you don't know it to 'invent' it?

P.S. How did you calculate the exact count of all problems involving medians, angle bisectors and the heights?