Turkey TST 1998 Problem 5, AF=AC

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Turkey TST 1998 Problem 5, AF=AC

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Source: Turkey TST 1998 Problem 5

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#1 h Dec 2, 2011, 12:26 AM • 1 Y

Y by Adventure10

In a triangle $ABC$ , the circle through $C$ touching $AB$ at $A$ and the circle through $B$ touching $AC$ at $A$ have diﬀerent radii and meet again at $D$ . Let $E$ be the point on the ray $AB$ such that $AB = BE$ . The circle through $A$ , $D$ , $E$ intersect the ray $CA$ again at $F$ . Prove that $AF = AC$ .

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#2 h Dec 2, 2011, 8:23 AM • 1 Y

Y by Adventure10

I will prove that if $F$ is the the point in ray $CA$ such that $AF = AC$ then $ADEF$ is cyclic.

Let $C_1$ be the circle through $C$ that touches $AB$ at $A$ and $C_2$ the circle through $B$ that touches $AC$ at $A$ .

Invert about $A$ . This sends $B$ and $C$ to $B'$ and $C'$ . Also, this inversion sends $C_1$ to the lines that passes through $C$ and is parallel to $AB$ and it sends $C_2$ to the lines through $B$ parallel to $AC$ . $D'$ is the intersection of this lines so $AB'C'D'$ is a parallelogram.

We see that $AE \cdot AE' = r^2$ so $AE' = \frac{r^2}{AE}$ .

Also, $B'E' = \frac{BE \cdot r^2}{AB \cdot AE} = \frac{r^2}{AE} = AE'$ so $E'$ is midpoint of $AB'$

Since $AF \cdot AF' =r^2 = AC \cdot AC'$ we have $AF' = AC'$ so $A$ is the midpoint of $C'F'$

Clearly, since $AC' \parallel B'D'$ we have $\angle F'AE' = \angle E'B'D'$ . Also $AF' = AC' = B'D'$ and $AE' = B'E'$ so triangles $F'AE'$ and $D'B'E'$ are congruent so $\angle F'E'A = \angle D'E'B'$ and therefore $D',E',F'$ are collinear. Hence, $ADEF$ is cyclic.

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#3 h Dec 3, 2011, 8:38 AM • 2 Y

Y by Adventure10, Mango247

Here is a simpler solution:-
Note that triangles $ADB$ and $ADC$ are similar because $\angle DAB=\angle ACD$ and $\angle ABD=\angle DAC$ .Now $CA$ meets $\odot ADE$ at $F$ .So $\angle DFA=\angle DEA\Longrightarrow \triangle DFC\sim \triangle ADE$ .
Hence $A\cap\triangle DFC\sim B\cap\triangle ADE$ .So we conclude $AF=AC$ .

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#4 h Aug 28, 2024, 5:52 PM

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By tangencies, we have
$\angle BAD=\angle DCA \quad \text{and}\quad \angle ABD=\angle DAC$ which implies $\triangle ABD\sim \triangle CAD$ . Thus,
$\dfrac{AB}{AC}=\dfrac{BD}{AD}\Longleftrightarrow AC=\dfrac{AB\cdot CD}{AD}$ On the other hand, observe that
$\angle CDF=\angle DCA-\angle DFC=\angle BAD-\angle AED=\angle ADE$ Therefore $\triangle AED\sim \triangle CFD$ . Thus,
$\dfrac{CF}{AE}=\dfrac{CD}{AD}\Longleftrightarrow CF=\dfrac{AE\cdot CD}{AD}$ To sum up
$AC=\dfrac{AB\cdot CD}{AD}=\dfrac{AE\cdot CD}{AD}=CF$ as desired.

This post has been edited 1 time. Last edited by ehuseyinyigit, Aug 28, 2024, 5:52 PM

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#5 h Dec 9, 2024, 2:47 PM

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$\angle BAD = \angle ACD, \angle ABD = \angle DAC \implies \triangle DAB \sim \triangle DCA$
$\angle BED = \angle DFA, \angle FDE = \angle FAE = \angle ADC \implies DCAF \sim DABE$ .
Since $B$ is midpoint of $AE$ , hence $A$ is midpoint of $FC$ and we are done. $\blacksquare$

This post has been edited 1 time. Last edited by andrewthenerd, Dec 9, 2024, 2:48 PM

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