SOLUTION: Can there exist a triangle ROS in which the trisectors of angle O intersect RS at D and E with RD =1,,DE=2,andES=4? Explain Thanks for any help .

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Question 1120393: Can there exist a triangle ROS in which the trisectors of angle O intersect RS at D and E with RD =1,,DE=2,andES=4? Explain
Thanks for any help .
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


No; it is not possible.

Consider first the fact that an angle bisector in a triangle divides the opposite side into two parts in the same ratio as the lengths of the two sides of the angle.

In triangle ROE, OD is the angle bisector. If RD=1 and DE=2, OE is twice the length of OR. Let OR=x and OE=2x.

In triangle DOS, OE is the angle bisector. If DE=2 and ES=4, OS is twice the length of OD. Let OD=y and OS=2y.

Now use Stewart's Theorem in triangles ROE and DOS to get two equations relating x and y.

Triangle ROE:
2x%5E2%2B4x%5E2+=+3y%5E2%2B6
3y%5E2+=+6x%5E2-6
y%5E2+=+2x%5E2-2

Triangle DOS:
4y%5E2%2B8y%5E2+=+24x%5E2%2B48
12y%5E2+=+24x%5E2%2B48
y%5E2+=+2x%5E2%2B4

But now we have two equations, based on the given information, that say in one case that y^2=2x^2-2 and in the other case y^2 = 2x^2+4.

So the given conditions lead us to two equations that are incompatible; that means the given conditions are not possible.

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.

Make a sketch.  I will refer to it as if you have it in front of you (as I have it in front of my eyes).


    +---------------------------------------------------------------+
    |    Let ROS be a triangle, and let its angle ROS at vertex O   |
    |    is trisected in a way that RD= 1, DE=2 and ES=4.           |    (*)
    +---------------------------------------------------------------+


I am going to bring it to contradiction and prove, in this way, that such trisection IS NOT POSSIBLE.



Let "t" be any of the three congruent small angles at vertex O.

Let "a" be the length of RO;  let "b" be the length of DO.


Then  the length of  EO  is  2a  (bisector theorem for triangle REO), 
and   the length of  SO  is  2b  (bisector theorem for triangle DOS).


    I will remind you, that the bisector theorem STATES that
    the ratio of the segments lengths, to which the bisector of a triangle
    divides the opposite side is equal to the ratio of the lengths
    of the two lateral corresponding sides.
    
    
Now,  triangle ROD has side lengths "a" and "b", that conclude the angle "t", and the opposite side RD= 1.

So,  we write the Cosine Law formula for RD

     RD^2 = a%5E2+%2B+b%5E2+-+2abcos%28t%29 = 1^2.                  (1)



Next, triangle DOE has side lengths "b" and "2a", that conclude the angle "t", and the opposite side DE= 2.

So, we write the Cosine Law formula for DE

    DE^2 = %282a%29%5E2+%2B+b%5E2+-+2%282a%29bcos%28t%29 = 2^2.            (2)



Finally, triangle EOS has side lengths "2a" and "2b", that conclude the angle "t", and the opposite side ES= 4.

So, we write the Cosine Law formula for ES

    ES^2 = %282a%29%5E2+%2B+%282b%29%5E2+-+2%282a%29%282b%29cos%28t%29 = 4^2.      (3)



I will rewrite the three equations (1), (2) (3) in simpler form

            a%5E2+%2B+b%5E2+-+2abcos%28t%29 = 1.                   (1')

            4a%5E2+%2B+b%5E2+-+4abcos%28t%29 = 4.                  (2')

            4a%5E2+%2B+4b%5E2+-+8abcos%28t%29 = 16.                (3')


Now compare equations  (1')  and  (3').   You see that the left side of (3') is four times the left side of (1'),

but the right side of (3') is sixteen times left side of (1').


It is the contradiction, and this contradiction proves that the original assumption (*) is not valid / (is not possible).

At this point, my proof is completed.


//////////////


It is also possible to construct another proof, using the similarity properties of small triangles. 

    Indeed, from one hand side, triangles ROD and EOS must be similar with the similarity coefficient of 2,

    because their "lateral" sides are proportional  RO : DO = a : b = EO : SO = (2a) : (2b) and the concluded angles "t" are congruent;


    but from the other hand side,  their sides, opposite to angle "t", are with the ratio 4:1, which gives the CONTRADICTION.