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A triangle has side lengths $7,$ $8$, and $9.$ There are exactly two lines that simultaneously bisect the perimeter and area of the triangle. Let $\theta$ be the acute angle between these two lines. Find $\tan \theta.$ 

 Aug 8, 2022
 #1
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Let the triangle be ABC

We can find the area of ABC thusly

Semi-perimeter  =   (9 + 8 + 7)  / 2    =  12

Area  =   sqrt [  12 (12 -9)(12 -8) (12-7) ] =  sqrt [ 12 * 3 * 4 * 5 ] =  12sqrt (5)

The height can  be figured as

12sqrt (5)   =  (1/2) (8) (h)

(3)sqrt (5)  = h

 

Line CD  divides the perimeter in half

Line CE divides the area in  half

CF  =  the height

 

Angle EDC is the acute angle (theta)  between  these two segments

 

 angle  EDC  =  angle DCF  - angle ECF

 

tan   DCF  = DF / CF  =  3 / ( 3sqrt (5))    =  1/sqrt (5)

tan  ECF  = Ef/ CF =   2 / (3sqrt (5) )

 

tan (EDC)  = tan [ DCF - ECF]  =   [  tan DCF - tan ECF ]  /  [ 1 + tan(DCF)* tan(ECF] ] =

 

tan (EDC)  =   [ 1/sqrt (5)  - 2/ (3sqrt 5 ) ]  /  [ 1 + (1/sqrt 5)(2/(3sqrt 5)) ] =

 

[ 1 / (3sqrt 5 ]  / [  1 + 2 / 15]  =

 

15  / [  17 * 3 sqrt 5 ]  =

 

15 / (51*sqrt 5 )  =

 

(5 / (17*sqrt 5) =

 

sqrt (5)  / 17

 

cool cool cool

 Aug 8, 2022

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