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Given a triangle ABC. For an arbitrary interior point X of the triangle denote by A1(X) the point intersection of the lines AX and BC, denote by B1(X) the point intersection of the lines BX and CA, and denote by C1(X) the point intersection of the lines CX and AB.

Determine all points P in the interior of the triangle for which each of the quadrilaterals AC1(P)PB1(P), BA1(P)PC1(P) and CB1(P)PA1(P) has an inscribed circle.

This is what the figure looks like

What I have been able to prove includes-

i) B3A1P = A4B1P, likewise for other corresponding pairs

ii) AB1P-AC1P = PB1P-PC1P=B2C1=DA1P , likewise for other corresponding pairs

iii) AC1P+BA1P+CB1P = s (semi perimeter of ∆ABC)

I know that this point must be the Inner Soddy Center of ∆ABC but am finding it difficult to prove it.

Please help..

Gerry Myerson
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user94529
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1 Answers1

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Finally cracked it!!

To show that point P is the inner soddy centre, we show that P lies on a hyperbola with focii $B,C$ and passing through $D$.

That means that we effectively have to show that $$CP-BP=CD-BD$$

Notice that $CB_1=CB_3=$ Length of Tangent from $C$ to the circle $O_B$.

Hence, $$\begin{align}CP+B_1P =CB_3 \\ \implies CP=CB_3-B_1P \end{align}$$ Similarly,$$BP=BC_4-PC_2$$ So, $$\begin{align} CP-BP &=CB_3-BC_4+(PC_2-PB_1) &(1)\\ &=(CC_4+B_3C_4)-(BB_3+B_3C_4)+(PC_1-PB_2) &(2) \\ &=CC_4-BB_3+B_2C_1 &(3) \\ &=CA_{1P}-BD+DA_{1P} &(4) \\ &=CD-BD &(5) \end{align}$$ Also, $$BP-AP=BF-AF$$ $$AP-CP=AE-CE$$

Therefore P lies on the intersection point of all three Hyperbolae that i mentioned earlier-This must be a unique point.

Now, the inner Soddy Centre satisfies these properties. Thus, it is the required point P.

Blue
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user94529
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  • I edited your formatting a bit for clarity. Please double-check that I didn't change the nature of your equations. – Blue Sep 18 '13 at 13:06
  • I get a little lost at step $(4)$. It might help if I knew exactly how $D$, $E$, $F$ are defined. Are they the feet of perpendiculars from $P$? (The diagram makes them look close, but not quite.) – Blue Sep 18 '13 at 13:11
  • $D, E, F$ are defined to be the feet of the perpendicular from the incentre $I$(the small grey point near $P$) of the triangle. Step $(4)$ follows from results that i was able to prove before i posted this question online. (They are the ones that i have written below the Figure in the question.) – user94529 Sep 18 '13 at 16:48
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    They can be derived by considering a general case wherein A triangle $ABC$ is divided into two triangles $ABX$ and $ACX$ by a line passing through $A$. Now, draw the incircles of triangles $ABX$ and $ACX$ and try proving that the length of the internal tangent of the circles is equal to $DX$ where $D$ is the foot of the perpendicular from the incircle upon $BC$. I think i should add that to my answer. :) – user94529 Sep 18 '13 at 16:59