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Source: China TST 2002 Quiz

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April

#1 h Mar 29, 2007, 8:48 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B.$ Points $C$ and $D$ are on circles $\omega_{1}$ and $\omega_{2},$ respectively,
such that lines $AC$ and $AD$ are tangent to circles $\omega_{2}$ and $\omega_{1},$ respectively.
Let $I_{1}$ and $I_{2}$ be the incenters of triangles $ABC$ and $ABD,$ respectively. Segments $I_{1}I_{2}$ and $AB$ intersect at $E$ .
Prove that: $\frac {1}{AE} = \frac {1}{AC} + \frac {1}{AD}$

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#2 h Mar 30, 2007, 5:46 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

The idea of the problem is how to use the incenters here. If you notice that $\measuredangle{ABC}= \measuredangle{ABD}$ and triangles $ABC$ and $DBA$ are similar, you are in the right track.

Proof

It is first easy to see that $\measuredangle{BAC}= \measuredangle{BDA}, \measuredangle{ACB}=\measuredangle{BAD}$ . Hence $\measuredangle{ABC}= \measuredangle{ABD}=\beta$ and $ABC$ and $DBA$ are similar.

It follows that $\measuredangle{I_{1}BI_{2}}= \beta$ and $\frac{BI_{1}}{BI_{2}}= \frac{BC}{BA}$ . Thus, $I_{1}BI_{2}, ABC, DBA$ are similar.

Let $M$ be the intersection of $BI_{1}$ and $AC$ . Then $\frac{BE}{BM}= \frac{BI_{1}}{BA}$ .

Since $BM = \frac{2BA.BC.cos \frac{\beta}{2}}{BA+BC}$ and $2.BI_{1}cos \frac{\beta}{2}= BC+BA-AC$ , we obtain $\frac{BE}{AB}= \frac{BA+BC-CA}{BC+BA}$ Thus $\frac{AE}{AB}= \frac{AC}{BA+BC}$ .

Therefore, $\frac{1}{AE}= (\frac{BC+BA}{AB}) . \frac{1}{AC}= \frac{1}{AC}+\frac{1}{AD}$ .

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yetti

#3 h Jan 29, 2009, 6:43 PM • 2 Y

Y by Adventure10, Mango247

Lemma: $H$ is orthocenter of a $\triangle ABC,$ the A-altitude $AH$ cuts the triangle circumcircle $(O)$ again at $X.$ $AH_1, AH_2$ are reflections of $AH$ in $AB, AC.$ Parallels through $H$ to $AH_2, AH_1$ cut $AH_1, AH_2$ at $K_1, K_2,$ forming a parallelogram $AK_1HK_2.$ Then $AK_1 + AK_2 = AX.$

Let $(O_1),\ X_1 \in (O_1)$ be reflections of $(O),\ X \in (O)$ in $AB.$ $\angle AHK_1 = \angle HAK_2 = 180^\circ - 2 \angle C$ $\Longrightarrow$ $\angle XHK_1 = 2 \angle C.$ As $\angle BHX = \angle HXB = \angle C,$ $HB$ bisects the angle $\angle XHK_1.$ Obviously, $AB$ bisects the angle $\angle HAK_1.$ Then $B$ is the A-excenter of the $\triangle AHK_1$ and $K_1B$ bisects the $\angle X_1K_1H.$ It follows that $X_1BHK_1$ with $X_1B = XB = HB$ is a kite and $K_1X_1 = HK_1.$ In conclusion, $AK_1 + AK_2 = AK_1 + HK_1 = AK_1 + K_1X_1 = AX_1 = AX.$

Problem: $(O), (P)$ are the 2 circles intersecting at $A, B.$ $(O), (P)$ cut the perpendicular bisector $OP$ of $AB$ at $M, N$ on the opposite sides of $AB$ than $C, D.$ Let $AB, AC, AD$ cut $OP$ at $K, U, V.$ Then $(O), (P)$ are A-Apollonius circles of the right $\triangle AKV, \triangle AKU$ $\Longrightarrow$ $AM, AN$ bisect the angles $\angle DAB, \angle CAB.$ Circles $(M), (N)$ with radii $MA, NA$ cut internal bisectors $AN, AM$ of the angles $\angle CAB, \angle DAB$ at the incenters $I, J$ of the $\triangle ABC, \triangle ABD.$ The circles $(M), (N)$ are centered on $AJ, AI,$ perpendicular to $AJ, AI.$ Inversion with center $A$ and power $AB^2$ takes $(O), (P)$ into parallels $o \parallel AD, p \parallel AC$ through $B$ to their tangents $AD, AC.$ If $C', D'$ are inversion images of $C, D,$ then $AC'BD'$ is a parallelogram, such that lines $AC', AD'$ are reflections of the line $AB$ in $AI, AJ.$ Lines $AI, AJ$ go to themselves and circles $(M), (N)$ to perpendiculars $m \perp AJ, n \perp AI$ through $B,$ intersecting $AI, AJ$ at the inversion images $I', J'$ of $I, J.$ Thus $B$ is orthocenter of $\triangle AI'J'.$ Line $IJ$ goes to its circumcircle $(Q)$ . $E$ goes to intersection $E'$ other than $A$ of the 3rd antitude $AB$ of the $\triangle AI'J'$ with its circumcircle $(Q).$ By the above lemma, $AE' = AC' + AD'$ $\Longrightarrow$ $\frac {AB^2}{AE} = \frac {AB^2}{AC} + \frac {AB^2}{AD}.$

EDIT: Hey, this was reply to a problem posted by shobber today and he did not mention any $\omega_2$ !!!

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Potla

#4 h Mar 18, 2011, 12:01 PM • 2 Y

Y by Adventure10, Mango247

Before reading this proof, one has to ensure that he is in a good mental state to read an unexpectedly large solution.
My proof runs as follows.
Firstly note that $AC$ and $AD$ are tangents to $\omega_2$ and $\omega_1,$ respectively. Therefore we obtain
$\begin{aligned}&\angle BCA+\angle CAB=\angle BAD+\angle ADB;\\&\implies \angle ABC=\angle ABD\implies \angle ABI_1=\angle ABI2.\end{aligned}$
Denote $G=BI_1\cap AC, F=BI_2\cap AD, X=BI_1\cap \omega_1, Y=BI_2\cap \omega_2.$
Since $\pi-\angle CAY=\angle YBA=\frac{\angle XBY}{2}=\angle XAG,$ therefore $X,A,Y$ are collinear. Now, note that $\begin{aligned}\angle GAF&=\pi-(\angle XAC+\angle YAD)\\&=\pi-2\cdot\frac{\angle XBY}{2}=\pi-\angle XBY;\end{aligned}$
So $AGBF$ is a cyclic quadrilateral. Since $\angle GBA=\angle FBA,$ therefore $\angle GAF=\angle GFA\implies AG=AF.$
Again, note that using this in accordance with the angle-bisector theorem, we obtain $\frac{AG}{AB}=\frac{AF}{AB}\implies \frac{BI_1}{I_1G}=\frac{BI_2}{I_2F},$ implying that $I_1I_2\parallel GF.$ Since $\angle GFB=\angle CAB=\angle AYB,$ therefore $XY\parallel GF,$ so that $I_1I_2\parallel GF\parallel XY.$
Using this we obtain that
$\begin{aligned}\angle I_1EA=\angle EAY&=\angle DAB+\angle DAY=\angle ACB+\angle DAY\\&=\angle ACB+\angle DBY=\angle ACB+\angle GBC\\&=\angle I_1GA.\end{aligned}$
Therefore $\triangle AGI_1\cong\triangle AEI_1\implies AG=AE=AF.$
On the other hand, we have $\frac{AF}{FD}=\frac{CB}{AB};$ or, $\frac{AF}{AD}=\frac{AB}{BD+AB}.$
By similar means we have its analogue $\frac{AG}{GC}=\frac{AB}{CB+AB},$ so the relation that we are required to prove is equivalent to with
$\frac{AG}{AC}+\frac{AF}{AD}=1, \iff \frac{AB}{CB+AB}+\frac{AB}{BD+AB};$
Which is equivalent to $AB^2=BC\cdot BD.$
This time, note that $\triangle BCG\sim\triangle BAX$ (Tuberculosis!) $\implies \boxed{BG\cdot BX=AB\cdot BC}.$ Analogously we obtain $\boxed{BF\cdot BY=AB\cdot BD}.$
Multiplying the last two relations leads to
$\boxed{AB^2\cdot BC\cdot BD=BG\cdot BX\cdot BF\cdot BY} \ \ \ \ \ (*).$
Now, note that
$\left.\begin{array}{ccc}\angle GBA&=\angle YBA\\ \angle GAB&=\angle AYB\end{array}\right\}\implies \triangle AGB\sim\triangle AYB;$
So that $\boxed{AB^2=BY\cdot BG}.$ From $\triangle ABX\sim\triangle ABF$ we obtain its analogue $\boxed{AB^2=BX\cdot BF}.$
Multiplying the last two relations together we get
$\boxed{BX\cdot BF\cdot BY\cdot BG=AB^4}. \ \ \ \ \ (**).$
Substituting the RHS in $(*)$ with RHS of $(**)$ we get
$AB^2\cdot BC\cdot BD=AB^4\implies AB^2=BC\cdot BD;$
As required! We are done. $\Box$

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#5 h Mar 18, 2011, 12:43 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

That is my proof.
Let $O_{1}$ , $O_{2}$ is the center of circumcircle of $ABC$ , $ABD$ .
$O_{1}O_{2}$ meets $AI_{1}$ , $AI_{2}$ at $M$ , $N$ . $BI_{1}$ meets $AC$ at $Q$ and $BI_{2}$ meets $AD$ at $P$ .
Easy to see that: $A$ , $M$ , $I_{1}$ is collinear and $A$ , $N$ , $I_{2}$ is collinear.
Have $\angle MI_{1}N=\angle MAN=\angle MI_{2}N$ hence $MNI_{1}I_{2}$ is inscribed. then $\angle AI_{1}E=\angle ANM=\frac{\pi}{2}-\angle ACB=\angle AI_{1}Q$ hence $AE=AQ$ similar $AE=AP$
Then have $\frac{AE}{AD}=\frac{AB}{AB+BD} and \frac{AE=AQ}{AC}= \frac{AB}{AB+BC}$ (*)
Note that $BC.BD=AB^{2}$ (**)
When (*).(**) we have QED.

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kunalsinghal1994

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kunalsinghal1994

#6 h Mar 21, 2011, 5:29 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Mine is an algebraic solution:
clearly triangles $ADB$ and $CAB$ are similar.
let $CA=r.AD$ where $r<1$ (assuming that CAB is smaller than ADB) and $AD=x.AB$
now we normalize $AB=1$
then we get
$AD=x$
$BD=1/r$
$BC=r$
$AC=rx$
so we have
$\frac{2}{\frac{1}{AC}+\frac{1}{AB}}=\frac{2rx}{1+r}$
thus we need to show that $2AE(1+r)=2rx$ .
drop perpendiculars from $I_1$ and $I_2$ on $AB$ at $P$ & $Q$ respectively.
so, $2PB=AB+BC-AC$ and $2PQ=AB+BD-AD$
thus, $PQ=(1-r)(1+\frac{1}{r}+x)$
also $I_1PE$ ~ $I_2QE$ with $PE=r.QE$
thus $2QE=(\frac{1-r}{1+r})(1+\frac{1}{r}+x)$
also $2AQ=AB+AD-BD=1+x-\frac{1}{r}$
and thus $2AE(1+r)=2rx$ as desired.

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