Triangle ABC has side lengths A B = 7 , B C = 8 , and C A = 9. Its incircle Γ meets sides BC, CA, and AB at D, E, F respectively. Let AD intersect Γ at a point P ≠ D . . The circle passing through A and P tangent to Γ intersects the circle passing through A and D tangent to Γ at a point K ≠ A .. Find K F K E . .

Question

Triangle ABC has side lengths   A  B  =  7  ,  B  C  =  8  ,  and   C  A  =  9. Its incircle   Γ meets sides BC, CA, and AB at D, E, F respectively. Let AD intersect   Γ at a point   P  ≠  D  . . The circle passing through A and P tangent to Γ intersects the circle passing through A and D tangent to   Γ at a point   K  ≠  A  .. Find                      K        F                    K        E              . .

prisililoraiyk · Accepted Answer

Step 1Let the circle through A and D tangent to BC be       ω    1   and the circle through A and P tangent to   Γ be       ω    2   . Then by radical axes on       ω    1   ,       ω    2   and   Γ , we see that AK, BC, and the common tangent of       ω    2   and   Γ are concurrent at a point, say X. Now since PEDF is a harmonic quadrilateral, we see that EF passes through X, so   X  E  ⋅  X  F  =  X      D    2    =  X  K  ⋅  X  A , and consequently AKFE is cyclic. Now let       ω    1   intersect AC and AB at Q and R respectively. Then by spiral similarity,   △  K  E  Q  ∼  △  K  F  R , so             K      F              K      E        =            F      R              E      Q       , and all there is left is to compute these two lengths. The side lengths of the triangle are   a  =  8 ,   b  =  9 and   c  =  7 , and the semiperimeter   s  =  12 . Since D, E, F are the touchpoints of the incircle with the sides, we have   B  D  =  B  F  =  s  −  b  =  12  −  9  =  3 , and   C  D  =  C  E  =  s  −  c  =  12  −  7  =  5 . Therefore,  E  Q  =  C  E  −  C  Q  =  C  D  −            C              D        2                    C      A        =  5  −      25    9    =      20    9  and similarly,  F  R  =  B  F  −  B  R  =  B  D  −            B              D        2                    B      A        =  3  −      9    7    =      12    7    . .Hence finally,            K      F              K      E        =            F      R              E      Q        =      12    7    ⋅      9    20    =      27    35    . .

mybabe43nncck · Accepted Answer

Step 1Find measure of   ∠  B  A  C  ≈      58.4    o   from the measures of the sides of triangle ABC.Find the measure of radius of circle   Γ from this relation:  r  =                    (        p        −        a        )        (        p        −        b        )        (        p        −        c        )            p      where   p  =            a      +      b      +      c        2   , you must get   r  ≈  2  ,  24Find AE it must be 4Extend KF to touch circle Gamma at point H. show that:  ∠  H  K  E  =  ∠  F  K  E  =≈  58  ,      4    o    K  H  =  K  E and   K  F  =  K  J , where J is intersection of circle Gamma with KE.we have   K  H  =  K  E  ≈  K  GDraw a tangent from K to circle   Γ to touch it at point I.Connect K to G ,   ∠  G  K  I  =      58.4    2    =      29.2    o   and you have r , find measure of   K  I  =  4 . Also calculate the measure of KG, it is about 4.4.Now use this relation:  (  K  F  =  K  J  )  ×  (  K  E  =  K  G  =  4.4  )  =  K      I    2  you get:            K      F              K      E        =      3.4    4.4    =≈  0.77

Triangle ABC has side lengths A B = 7 , B C = 8 , and C A

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