+0  
 
0
59
1
avatar

In triangle ABC, AB = 14, AC = 13, and BC = 15.  Let X be the reflection of A over B, and let CX intersect the circumcircle of triangle ABC again at D.  Find the length CD.

 

 Feb 21, 2020
 #1
avatar+109450 
+1

[ Because  of  the nasty roots that we will find in this problem....I leaned heavily on WolframAlpha for  most of the computations ] 

 

Note that   AB    = BX   so.....AX   = 28

 

Using the Law of Cosines    we have

BC^2  = AB^2  + AC^2 - 2(AC*AB) cos CAX

15^2  =  14^2  +13^2  - 2(13 * 14) cos CAX

Solving for the cos CAX  we  get  that  cos (CAX)  = 5/13

So....the sin of CAX =  12/13 

 

And we can find   CX  using the Law of Cosines  again

CX^2  = AC^2  + AX^2  - 2(AC * AX) cos CAX

CX^2  = 13^2  + 28^2  - 2(13 * 28) (5/13)

Solving this for  CX  gives us  CX  =   √673

 

And again  we can find the cosine of angle  BCX  as

BX^2  =  BC^2  + CX^2  - 2(BC * CX) cos BCX

14^2  = 15^2  + 673  -2 ( 15 * √673) cosBCX 

Solving this  for  cos BCX gives  us   117/ [ 5 √673 ]

 

And sin BCX   =  √  [  1   - [ 117/ [ 5 √673 ] ]^2 ]    =   56 / [ 5√673]

 

Connect  BD

 

Since  quadrilateral ABDC  is inscribed in a circle....angles  CAB  and   CDB are supplemental....so the have the same sines  =12/13

 

So.....using the Law of Sines

BC / sin  CDB  =   BD  /sin BCX

15 / (12/13)  = BD  / [  56 / [ 5√673] ] 

Solving this gives that  BD   =  182  / √673

 

 

Finally.....using the Law of Cosines we  can  find   CD  as folllows

 

BD^2  =   BC^2  + CD^2   -  2 (BC * CD) cos BCX

Let  CD   = x   and we have that

182^2 / 673   =   15^2 + x^2  -  2 (15 * x)*(117/ [5sqrt (673)] )

 

Solving this for x  = CD  gives us two  values

 

CD  = 281 / √673  ≈  10.83   or      

 

CD  =  421 / √673  ≈  16.23

 

But     BC  =  15   and is clearly  greater than CD

 

So   CD =   281 / √673   ≈  10.83   is  the  correct solution

 

 

cool cool cool

 Feb 21, 2020
edited by CPhill  Feb 21, 2020

26 Online Users

avatar
avatar