fixed point for a circle, tangent to a given line

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fixed point for a circle, tangent to a given line

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Source: 2016 Oral Moscow Geometry Olympiad grades 10-11 p5

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#1 h Jul 29, 2019, 8:43 AM • 2 Y

Y by Adventure10, Mango247

From point $A$ to circle $\omega$ tangent $AD$ and arbitrary a secant intersecting a circle at points $B$ and $C$ (B lies between points $A$ and $C$ ). Prove that the circle passing through points $C$ and $D$ and touching the straight line $BD$ , passes through a fixed point (other than $D$ ).

This post has been edited 1 time. Last edited by parmenides51, Jul 29, 2019, 3:00 PM
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#2 h Jul 29, 2019, 11:57 AM • 1 Y

Y by Adventure10

inversion at D,and the constant point is the reflection of A' wrt ω'

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#3 h Dec 30, 2019, 11:12 AM • 4 Y

Y by GeoMetrix, strawberry_circle, Adventure10, Mango247

2016 Oral Moscow Geometry Olympiad P5 wrote:

From point $A$ to circle $\omega$ tangent $AD$ and arbitrary a secant intersecting a circle at points $B$ and $C$ (B lies between points $A$ and $C$ ). Prove that the circle passing through points $C$ and $D$ and touching the straight line $BD$ , passes through a fixed point (other than $D$ ).

After Inversion at $D$ with an arbitary radius the problem gets equivalent to this problem:-

Inverted Problem wrote:

A line contains points $A',D$ and a line $\|$ to $A'D$ be named as $\omega'$ and let $B'$ be an arbitary point on $\omega'$ and let $\odot(B'A'D)\cap\omega'=C'$ . Let a line $\|$ to $B'D'$ through $C'$ intersect $A'D$ at $T'$ . Then $C'T'$ passes through a fixed point other than $C'$

Let $T'C'\cap\odot(DA'B'C')=K.$ . Then $\angle A'DB'=\angle A'CB'=\angle A'T'K'=\angle B'C'K'\implies \text{Reflection of } A' \text{ over } \omega' \text{ lies on } C'T' \text{ which is the fixed point}$
Hence, Inverting back we get that Midpoint of $DE$ is the fixed point, where $E$ is the Point of Tangency made by the second tangent from $A$ to $\omega$ . $\qquad\blacksquare$

This post has been edited 12 times. Last edited by amar_04, Dec 30, 2019, 1:08 PM

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#4 h Dec 30, 2019, 12:05 PM • 2 Y

Y by Adventure10, Mango247

parmenides51 wrote:

From point $A$ to circle $\omega$ tangent $AD$ and arbitrary a secant intersecting a circle at points $B$ and $C$ (B lies between points $A$ and $C$ ). Prove that the circle passing through points $C$ and $D$ and touching the straight line $BD$ , passes through a fixed point (other than $D$ ).

Let $E$ be the other point of tangency from $A$ to $\omega$ .Thus $DBCE$ is harmonic.Let $M$ be the midpoint of $DE$ .Thus $\measuredangle{MCD}=\measuredangle{ECA}=\measuredangle{ECB}=\measuredangle{EDB}=\measuredangle{MDB}$ .Thus $\odot{MDC}$ is tangent to $BD$ and since $M$ is fixed for a particular $A$ so is the required point. $\square$

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