Bad Karma is a Consequence of Guessing

Geometry Level 4

A B C \triangle ABC contains four inscribed circles. Seqments J I , K L , JI, KL, and M H MH are parallel to the sides of A B C \triangle ABC which they do not intersect. Compare the radius of circle X X to the sum of the radii of circles Y Y , Z Z , and W W .

X = Y + Z + W X = Y + Z + W X > Y + Z + W X > Y + Z + W X < Y + Z + W X < Y + Z + W

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2 solutions

Sathvik Acharya
Jan 26, 2021

Since J I A B JI\parallel AB , H M C A \;\;HM\parallel CA\; and K L C B \;\;KL\parallel CB , A B C J I C M B H A L K \triangle ABC\sim \triangle JIC\sim\triangle MBH\sim \triangle ALK M B A B = W X , J I A B = Y X , A L A B = Z X (1) \begin{aligned} \implies \frac{MB}{AB}=\frac{W}{X} ,\;\; \frac{JI}{AB}=\frac{Y}{X} ,\;\; \frac{AL}{AB}=\frac{Z}{X} \;\; \end{aligned}\tag{1} Observe, A B C J I C M F L \triangle ABC\sim\triangle JIC\sim \triangle MFL Also, J I C \triangle JIC and M F L \triangle MFL have the same exradius ( X ) (X) , so we have, J I C M F L J I = L M (2) \triangle JIC\cong \triangle MFL\implies JI=LM \tag{2} Using ( 1 ) (1) and ( 2 ) (2) , M B A B + J I A B + A L A B = W X + Y X + Z X M B + L M + A L A B = W + Y + Z X X = W + Y + Z \begin{aligned} \frac{MB}{AB}+\frac{JI}{AB}+\frac{AL}{AB}&=\frac{W}{X}+\frac{Y}{X}+\frac{Z}{X} \\ \\ \implies \frac{MB+LM+AL}{AB}&=\frac{W+Y+Z}{X} \\ \\ \therefore \boxed{X=W+Y+Z} \end{aligned}

Well done!

Fletcher Mattox - 4 months, 2 weeks ago

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Thank you!

Sathvik Acharya - 4 months, 2 weeks ago
Mark Hennings
Jan 26, 2021

The triangles A B C ABC , A L K ALK , M B H MBH and J I C JIC are all similar, and so we deduce that Z = h a 2 X h a X W = h b 2 X h b X Y = h c 2 X h c X Z \; = \; \frac{h_a-2X}{h_a}X \hspace{1cm} W \; =\; \frac{h_b - 2X}{h_b}X \hspace{1cm} Y \; = \; \frac{h_c - 2X}{h_c}X where h a , h b , h c h_a,h_b,h_c are the lengths of the altitudes of the triangle A B C ABC from the vertices A , B , C A,B,C respectively. This is because the length of the altitude from A A in the triangle A L K ALK is h a 2 X h_a - 2X , and so on. Thus Y + Z + W = 3 X 2 X 2 ( h a 1 + h b 1 + h c 1 ) = 3 X X 2 Δ ( a + b + c ) = 3 X 2 X 2 s Δ = 3 X 2 X = X \begin{aligned} Y + Z + W & = \; 3X - 2X^2\big(h_a^{-1} + h_b^{-1} + h_c^{-1}\big) \; = \; 3X - \frac{X^2}{\Delta}(a + b + c) \\ & = \; 3X - \frac{2X^2s}{\Delta} \; = \; 3X - 2X \; = \; X \end{aligned} where Δ , s \Delta,s are the area and semiperimeter of the triangle A B C ABC .

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