China 2010 quiz3 problem 1

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China 2010 quiz3 problem 1

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agzqu

#1 h Sep 12, 2010, 12:08 AM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a convex quadrilateral with $A,B,C,D$ concyclic. Assume $\angle ADC$ is acute and $\frac{AB}{BC}=\frac{DA}{CD}$ . Let $\Gamma$ be a circle through $A$ and $D$ , tangent to $AB$ , and let $E$ be a point on $\Gamma$ and inside $ABCD$ .
Prove that $AE\perp EC$ if and only if $\frac{AE}{AB}-\frac{ED}{AD}=1$ .

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#2 h Sep 12, 2010, 8:00 AM • 4 Y

Y by tenplusten, Adventure10, Mango247, and 1 other user

Very nice problem!

Let $M$ be the midpoint of $AC$ . Since $ABCD$ is harmonic, $BD$ is the symmedian of $\triangle ABC,\triangle ADC$ , so $\angle ABD=\angle CBM,\angle ADB=\angle CDM$ , so $\angle BMA=\angle MBC+\angle MCB=\angle ABD+\angle BCM=\angle ACD+\angle BCM=\angle BCD=\angle AMD$ . Note that $\angle AED=180-\angle DAE-\angle ADE=180-\angle DAB=\angle BCD=\angle AMD$ , so $DAME$ is cyclic. Now let $F$ be on the extension of $DE$ such that $\angle MFD=\angle ADM=\angle BDC=\angle BAM$ . Since $\angle ABM=\angle DAM=\angle MEF$ , we get $\triangle MBA\sim\triangle MAD\sim\triangle MEF$ , so by spiral symmetry, $\triangle BAE\sim \triangle ADF$ , so $\frac{BA}{AE}=\frac{AD}{DF}$ . Thus $\frac{AE}{AB}=\frac{DE+DA}{DA}\iff AD=EF\iff AM=ME\iff \angle AEC=90$ and we are done.

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cwein3

#3 h Nov 22, 2011, 6:25 AM • 2 Y

Y by Adventure10, Mango247

I will prove the only if part, the if follows by reversing the steps. Let $\gamma$ intersect $AC$ at $M$ and $DM$ intersect $AE$ at $X$ .

Lemma 1: $M$ is the midpoint of $AC$ .
Proof: Since $ABCD$ is harmonic, $DB$ is a symmedian of triangle $ADC$ . Let $M'$ be the midpoint of $AC$ , then we have $\angle DM'C = \angle DAB$ . But from the fact that $\gamma$ is tangent to $AB$ , we get $\angle DMC = \angle DAB$ , thus $M$ and $M'$ are the same point.

Now note that $ME = AM$ since $M$ is the midpoint of $AC$ . Thus, $DX$ bisects $\angle ADE$ .

Lemma 2: $AX = AB$ .
Proof: We have $\angle MAE = \angle MDA$ , so $\dfrac {MA} {MD} = \dfrac {AX} {DA}$ . Furthermore, a well known spiral similarity around $D$ maps $\triangle BAD$ to $\triangle CMD$ , so $\dfrac {AB} {DA} = \dfrac {MC} {MD} = \dfrac {MA} {MD}$ . Therefore, we get $AX = AB$ .

The rest is just algebra. From the angle bisector theorem, we have $\dfrac {AE} {AD + DE} = \dfrac {AX} {AD} = \dfrac {AB} {AD}$ . Now rearranging, we get the desired equation.

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skytin

#4 h Nov 24, 2011, 3:18 PM • 2 Y

Y by Adventure10, Mango247

Solution :
Easy to see that tangents from A , C to (ABCD) intersects at point K on DB
Let point A' is on ED , such that AE = EA' , E is on Г , so angle AEA' = BCD
let point C' is on (KCD) such that C'K = KC
Easy to see that C' is on BC , angle CDC' = 2*CDK , angle BAC' = 90
AE/EA' = CB/DC = AB/DA , AE/AB = EA'/DA = 1 + ED/DA . done

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#5 h Nov 27, 2011, 5:59 AM • 2 Y

Y by Adventure10, Mango247

Assume $M$ to be the mid point of $AC$ . Extend $DE$ to $P$ with $\angle APD=\angle AEB$ .
Hence $\triangle ABE \sim \triangle DAP$ , given $E$ is on a circle passes $A,D$ and tangent to $AB$ . Then $\frac{AE}{AB}=\frac{DP}{DA}$ .
$\angle BMA=\angle DMA=\angle BCD$ since quad $ABCD$ is harmonic(China National Maths Contest for Senior 2011 introduced this Lemma), then $\angle MAB=\angle MDA$ , $\angle MBA=\angle MAD$ , hence $M$ is in similar position in triangles $\triangle ABE,\triangle DAP$ .
Easily get $\triangle MAD \sim \triangle MEF$ , so $AE \perp EC$ equals $EM=MA$ equals $EF=AD$ equals $\frac{AE}{AB}-\frac{ED}{AD}=1$ .

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#6 h Aug 20, 2015, 3:10 AM • 2 Y

Y by Adventure10, Mango247

Let $\mathcal{I} : X \mapsto X'$ be the inversion with pole $A$ and arbitrary radius $r.$ Because $A, B, C, D$ are concyclic, it follows that $B', C', D'$ are collinear. Moreover, since inversion preserves angles between intersecting curves, $\Gamma'$ is the line through $D'$ parallel to $AB'.$

From the inversive distance formula, we find $\frac{B'C'}{D'C'} = \frac{BC \cdot \frac{r^2}{AB \cdot AC}}{DC \cdot \frac{r^2}{AD \cdot AC}} = \frac{BC}{AB} \cdot \frac{AD}{DC} = 1.$ It follows that $C'$ is the midpoint of $\overline{B'D'}.$ Let $P'$ be the reflection of $A$ in $C'.$ Note that the diagonals of quadrilateral $AB'P'D'$ bisect one another, which implies that $AB'P'D'$ is a parallelogram. Hence, $P' \in \Gamma'$ and $D'P' = AB'.$

Now, because $AE \cdot AE' = AB \cdot AB' = r^2$ , it follows that $AE : AB = AB' : AE'.$ Meanwhile, from the inversive distance formula, we obtain $E'D' = ED \cdot \frac{r^2}{AE \cdot AD} \implies \frac{ED}{AD} = E'D' \cdot \frac{AE}{r^2} = \frac{E'D'}{AE'}.$ Therefore, $\tfrac{AE}{AB} - \tfrac{ED}{AD} = 1 \iff AE' = AB' - E'D'.$ Note that $AB' - E'D' = P'D' - E'D' = P'E'.$ Since $C'$ is the midpoint of $\overline{AP'}$ it follows that $P'E' = A'E' \iff E'C' \perp A'C'$ , which is equivalent to $AE \perp CE$ , as desired. $\square$

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Arefe

#7 h Apr 8, 2020, 5:53 PM

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For the iff part , Let X be the intersection of AE with bisector of ADE , using the equation of the problem we can see that AB is equel to AX . Then we find that triangle DEX is similar to triangle DCB . Because of that triangle DEC is similar to triangle DXB and because the triangle XAB is isosceles we find that AEC = 90 and we are done .

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