2018 Hong Kong TST 2 Problem 1

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2018 Hong Kong TST 2 Problem 1

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Source: 2018HKTST2P1

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133 posts

YanYau

#1 h Oct 21, 2017, 11:44 AM • 4 Y

Y by falantrng, Davi-8191, Adventure10, Mango247

Let $ABC$ be a triangle with $AB=AC$ . A circle $\Gamma$ lies outside triangle $ABC$ and is tangent to line $AC$ at $C$ . Point $D$ lies on $\Gamma$ such that the circumcircle of triangle $ABD$ is internally tangent to $\Gamma$ . Segment $AD$ meets $\Gamma$ secondly at $E$ . Prove that $BE$ is tangent to $\Gamma$

This post has been edited 1 time. Last edited by YanYau, Aug 22, 2018, 2:25 AM

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1008 posts

#2 h Oct 21, 2017, 1:43 PM • 2 Y

Y by falantrng, Adventure10

My solution:
Suppose the tangent to $\Gamma$ and $(ABD)$ intersect $AC$ at $L.$
We know $AB^2=AC^2=AD\cdot AE\to \angle ABD=\angle BED.$ $(1)$
We know $\angle BEC=\angle BED-\angle CED=^{(1)} \angle ABD-\angle LDC=180-\angle BAD-\angle BDA-\angle LDC$ $=180-\angle BDL-\angle LDC-\angle BDA=180-\angle CDA=\angle CDE.$ As desired.

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133 posts

YanYau

#3 h Oct 21, 2017, 1:59 PM • 2 Y

Y by Adventure10, Mango247

This was my solution during the test, I solved it in the first 10 minutes:

By Power of a Point, $AC^2=AD\cdot AE$ .

Invert about $A$ with radius $r=AB=AC$ .

$B$ , $C$ and $\Gamma$ are invariant under the inversion, while $D$ maps to $E$ and $E$ maps to $D$ .

The image of circle $ABD$ will be a line passing through the images of $B$ and $D$ and is tangent to $\Gamma$ .

The image of $B$ , $D$ is $B$ , $E$ . Thus the circle maps to line $BE$ , and thus $BE$ is tangent to $\Gamma$

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308 posts

#4 h Oct 21, 2017, 2:27 PM • 1 Y

Y by Adventure10

Let the tangent at $E$ intersect circle $ABD$ at $B',K$ with $B'$ on the opposite side of $AC$ to $D$ . We will prove $AB'=AC \implies B=B'$ which completes the problem.

By a well-known lemma (similar triangles) we have $AB'^2=AE \cdot AD=AC^2 \implies AB'=AC$ as desired.

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#5 h Mar 8, 2018, 12:54 PM • 2 Y

Y by Adventure10, Mango247

Ferid.---. wrote:

My solution:
Suppose the tangent to $\Gamma$ and $(ABD)$ intersect $AC$ at $L.$
We know $AB^2=AC^2=AD\cdot AE\to \angle ABD=\angle BED.$ $(1)$
We know $\angle BEC=\angle BED-\angle CED=^{(1)} \angle ABD-\angle LDC=180-\angle BAD-\angle BDA-\angle LDC$ $=180-\angle BDL-\angle LDC-\angle BDA=180-\angle CDA=\angle CDE.$ As desired.

Is there some mistakes?

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8 posts

#6 h Mar 18, 2018, 3:02 AM • 2 Y

Y by Adventure10, Mango247

My students solution:
$BD$ meet $\Gamma$ at another point L
It is easy to show that $EL$ $//$ $AB$ ,then $\angle ABE = \angle BEL$
$AB^2=AC^2=AD\cdot AE\to \angle ABE=\angle ADB$ ,
then $\angle BEL=\angle EDL$ ,DONE.

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39 posts

#8 h Oct 25, 2020, 1:28 PM

Y by

How can we draw the diagram?

This post has been edited 1 time. Last edited by mathematicsy, Oct 25, 2020, 1:29 PM
Reason: grammar

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90 posts

#9 h Oct 25, 2020, 8:50 PM • 1 Y

Y by Math-48

Let $G=AC\cap (ABD) , E=AD\cap \Gamma , F=DG\cap \Gamma ,\quad \text{and} \quad H=DC\cap (ABD)$
Claim: $EF\parallel AG$
Proof. by Star Lemma , $HD$ bisects $\angle{ADG}$ so $CF=CE$ and the result follows.
$\blacksquare$
Now $AE.AD=\text{Pow}_\Gamma(A)=AC^2=AB^2$ so $\Delta ABE\sim \Delta ABD$
$\implies \angle{AEB}=\angle{ABD}=\angle{AGD}=\angle{EFD}.$
so we're clear :clap:

:clap:

@above you could draw the two circles firstly

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#10 h Oct 1, 2021, 4:23 PM • 1 Y

Y by teomihai

As $\Gamma$ is tangent to $\overline{AC}$ , then $AB^2=AC^2=AD\cdot AE$ , which means that $\overline{AB}$ is tangent to $(BED)$ . Now we do angle chase,
$\begin{align*} \measuredangle DEB=\measuredangle DBA=\measuredangle DCE, \end{align*}$ where the last equality comes from the fact that $(ABD)$ and $(CED)$ are tangent to each other. Done.

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1514 posts

#11 h Apr 18, 2022, 9:17 PM

Y by

YanYau wrote:

Let $ABC$ be a triangle with $AB=AC$ . A circle $\Gamma$ lies outside triangle $ABC$ and is tangent to line $AC$ at $C$ . Point $D$ lies on $\Gamma$ such that the circumcircle of triangle $ABD$ is internally tangent to $\Gamma$ . Segment $AD$ meets $\Gamma$ secondly at $E$ . Prove that $BE$ is tangent to $\Gamma$

Let $BD \cap \Gamma=F$ by PoP $AE \cdot AD=AC^2=AB^2$ so $AB$ is tangent to $BED$ now becuase $(ABD)$ and $\Gamma$ are tangent at $D$ we have $AB \parallel EF$ so by angle chase
$\angle EDF=\angle ABE=\angle BEF \implies BD \; \text{is tangent to} \; \Gamma$ Thus we are done

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