2022-03-11

Two circles are internally tangent at point $C$. $A$ is a point on outer circle and $AB,AD$ are two tangents to the inner circle, where $B,D$ are on the outer circle, and $E,F$ are the points of tangency on the inner circle. $EF,AC$ intersect at $G$. Prove: $\mathrm{\angle }EGB=\mathrm{\angle }FGD$.

Prove 1:

Let the two circles be $c_1,c_2$. $AC$ and $c_1$ intersect at $Q$. The tangent at $C$ intersect with extended $EF$ at $P$.

$AB,AD$ are tangents to $c_1⟹$ $EF$ is polar of $A$ to $c_1$. $P$ is on $EF$, according to La Hire Theorem, $A$ is on the polar of $P$ to $c_1$. That means $AC$ is the polar of $P$ to $c_1$. So the other point of tangency is $Q$ and $PQ$ is the tangent. Let the intersects of $PQ$ with $AB,AD$ be $M,N$ respectively.

Suppose $AB=b,AD=a,AE=AF=c$. Make circle $C_3$ centered at $A$ with radius $c$. Easy to see that $c_1,c_3$ are orthogonal circles, so the inversion of $c_1$ to $c_3$ is $c_1$ itself, then we know the inversion of $C$ to $c_3$ is $Q$.

$c_2$ passes through the center $A$ of $c_3$, so the inversion of $c_2$ to $c_3$ is the line passing the intersects of $c_2$ and $c_3$. Since $C$ is the point of tangency of $c_1$ and $c_2$, then the inversion of $c_1$ must be the tangent to $c_2$ at $Q$, i.e., the line $PQ$. Suppose the intersects of $PQ$ with $AB,AD$ are $M,N$ respectively. So

$BE=b-c,AM={\displaystyle \frac{A{E}^2}{AB}}={\displaystyle \frac{c^2}{b}}⟹EM=c-{\displaystyle \frac{c^2}{b}}$

$DF=a-c,AN={\displaystyle \frac{A{F}^2}{AD}}={\displaystyle \frac{c^2}{a}}⟹FN=c-{\displaystyle \frac{c^2}{a}}$

Apply Menelaus’ Theorem to $\mathrm{△}AEF$ and line $PM$:

${\displaystyle \frac{EM}{MA}}{\displaystyle \frac{AN}{NF}}{\displaystyle \frac{FP}{EP}}=1⟹{\displaystyle \frac{EP}{FP}}={\displaystyle \frac{c-{\displaystyle \frac{c^2}{b}}}{{\displaystyle \frac{c^2}{b}}}}{\displaystyle \frac{{\displaystyle \frac{c^2}{a}}}{c-{\displaystyle \frac{c^2}{a}}}}={\displaystyle \frac{b-c}{a-c}}$

At mean time $DCEQ$ is cyclic harmonic quadrilateral，and $P$ is a point outside $c_1$，$G$ is the intersect of $EF$ and the polar of $P$ to $c_1$，from the property of harmonic quadrilateral we know：

$(E,F;P,G)=-1⟹{\displaystyle \frac{EP}{FP}}÷{\displaystyle \frac{EG}{FG}}=1⟹{\displaystyle \frac{EG}{FG}}={\displaystyle \frac{EP}{FP}}={\displaystyle \frac{b-c}{a-c}}={\displaystyle \frac{EP}{FP}}$

So ${\displaystyle \frac{BE}{DF}}={\displaystyle \frac{EG}{FG}},AE=AF⟹\mathrm{\angle }BEG=\mathrm{\angle }DFG⟹\mathrm{△}BEG\sim \mathrm{△}DFG⟹\mathrm{\angle }EGB=\mathrm{\angle }FGD◼$

References

• Harmonic Geometry
• Poles and Polars
• Chapter 9 Projective Geometry from EGMO by Evan Chen

Chinese Version:

内外两个大小不等的圆内相切于点$C$。 $A$为外圆上一点，$AB,AD$为内圆两条切线，其中$B,D$在外圆上，$E,F$为内圆上对应的切点。$EF$与$AC$相交于$G$。证明：$\mathrm{\angle }EGB=\mathrm{\angle }FGD$。

证明:

设内外二圆为$c_1,c_2$。$AC$与$c_1$另一交点为$Q$，过$C$的外圆切线与EF延长线相交于$P$。

$AB,AD$为内圆切线，则$EF$所在直线为点$A$关于内圆$c_1$的极线。而$P$在$EF$所在直线上，根据 La Hire定理，$A$也在点$P$关于$c_1$的极线上，那么$AC$所在直线即为该极线，那么$PQ$为点$P$关于$c_1$的另一切线，$Q$即为相应切点。设该切线与$AB,AD$分别相交于$M,N$。

设$AB=b,AD=a,AE=AF=c$, 以$A$为圆、$c$ 为半径做圆$c_3$，显然可见，$c_1,c_3$是正交圆，根据反演性质，$c_1$关于$c_3$的反演为$c_1$本身，那么可知点$C$关于$c_3$的反演即为$Q$。

外圆$c_2$经过$c_3$的圆心$A$，则$c_2$关于$c_3$的反演为经过$c_2,c_3$二交点的直线。而$c_1,c_2$相切于$C$，根据反演性质，$c_1,c_2$关于$c_3$的反演也应该相切于$Q$，那么可见$c_2$关于$c_3$的反演即为经过$Q$点的关于$c_1$的切线，即$PQ$所在直线，则$PQ$与$AB,AD$的交点$M,N$即为$B,D$两点关于$c_3$的反演点。那么

$BE=b-c,AM={\displaystyle \frac{A{E}^2}{AB}}={\displaystyle \frac{c^2}{b}}⟹EM=c-{\displaystyle \frac{c^2}{b}}$

$DF=a-c,AN={\displaystyle \frac{A{F}^2}{AD}}={\displaystyle \frac{c^2}{a}}⟹FN=c-{\displaystyle \frac{c^2}{a}}$

对$\mathrm{△}AEF$及直线$PM$运用Menelaus定理：

${\displaystyle \frac{EM}{MA}}{\displaystyle \frac{AN}{NF}}{\displaystyle \frac{FP}{EP}}=1⟹{\displaystyle \frac{EP}{FP}}={\displaystyle \frac{c-{\displaystyle \frac{c^2}{b}}}{{\displaystyle \frac{c^2}{b}}}}{\displaystyle \frac{{\displaystyle \frac{c^2}{a}}}{c-{\displaystyle \frac{c^2}{a}}}}={\displaystyle \frac{b-c}{a-c}}$

同时，$DCEQ$为共圆$c_1$的调和四边形，$P$为$c_1$外点，$G$为$P$关于$c_1$的极线与$EF$的交点，根据调和四边形性质可知：

$(E,F;P,G)=-1⟹{\displaystyle \frac{EP}{FP}}÷{\displaystyle \frac{EG}{FG}}=1⟹{\displaystyle \frac{EG}{FG}}={\displaystyle \frac{EP}{FP}}={\displaystyle \frac{b-c}{a-c}}={\displaystyle \frac{EP}{FP}}$

故 ${\displaystyle \frac{BE}{DF}}={\displaystyle \frac{EG}{FG}},AE=AF⟹\mathrm{\angle }BEG=\mathrm{\angle }DFG⟹\mathrm{△}BEG\sim \mathrm{△}DFG⟹\mathrm{\angle }EGB=\mathrm{\angle }FGD◼$

Prove 2:

Let the two circles be $c_1,c_2$ and their centers are $O_1,O_2$ respectively.

Easy to know that $AE=AF,\mathrm{\angle }AEG=\mathrm{\angle }AFG$. From the Law of Sines:

${\displaystyle \frac{AG}{sin\mathrm{\angle }AEG}}={\displaystyle \frac{EG}{sin\mathrm{\angle }EAG}}={\displaystyle \frac{FG}{sin\mathrm{\angle }FAG}}⟹{\displaystyle \frac{EG}{FG}}={\displaystyle \frac{sin\mathrm{\angle }EAG}{sin\mathrm{\angle }FAG}}={\displaystyle \frac{sin\mathrm{\angle }BAC}{sin\mathrm{\angle }CAD}}={\displaystyle \frac{sin\mathrm{\angle }BDC}{sin\mathrm{\angle }CBD}}={\displaystyle \frac{BC}{BD}}$

Now we prove that $CE,CF$ bisect $\mathrm{\angle }ACB,\mathrm{\angle }ACD$ respectively:

Extended $CE$ and $c_2$ intersect at $H$. Apparently

$C{O}_2=H{O}_2,C{O}_1=E{O}_1⟹\mathrm{\angle }CE{O}_1=\mathrm{\angle }EC{O}_1=\mathrm{\angle }CH{O}_2⟹E{O}_1\parallel H{O}_2$

So $E{O}_1\perp AB⟹H{O}_2\perp AB⟹\stackrel{⌢}{AH}=\stackrel{⌢}{HB}⟹\mathrm{\angle }BCE=\mathrm{\angle }ACE$, similarly $\mathrm{\angle }ACF=\mathrm{\angle }DCF$

$⟹{\displaystyle \frac{AC}{AE}}={\displaystyle \frac{BC}{BE}}={\displaystyle \frac{AC}{AF}}={\displaystyle \frac{CD}{DF}}⟹{\displaystyle \frac{BC}{BE}}={\displaystyle \frac{CD}{DF}}⟹{\displaystyle \frac{BC}{CD}}={\displaystyle \frac{BE}{DF}}={\displaystyle \frac{EG}{FG}}$

$\mathrm{\angle }BEG=\mathrm{\angle }DFG⟹\mathrm{△}BEG\sim \mathrm{△}DFG⟹\mathrm{\angle }EGB=\mathrm{\angle }FGD◼$

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03/19/2022

In $\mathrm{△}ABC$, $\mathrm{\angle }A={24}^{\circ },\mathrm{\angle }C={30}^{\circ }$, $D$ is on $AC$ so that $AB=CD$. Prove: $AB\cdot BC=AC\cdot BD$.

Prove:

Make the mirror of $A$ along $BC$ to be $E$ so that $\mathrm{△}ABC\cong \mathrm{△}EBC$.

Easy to see that $AB=BE,AC=EC$ and $\mathrm{△}ACE$ is equilateral so $AC=AE,\mathrm{\angle }BEF={60}^{\circ }-\mathrm{\angle }BEC={36}^{\circ }$

Make $F$ on $AE$ so that $DF\parallel CE$ easy to know that $\mathrm{△}ADF$ is also equilateral, so

$AD=AF=DF⟹CD=AC-AD=AE-AF=EF=AB=BE$

$⟹\mathrm{\angle }EFB=\mathrm{\angle }EBF={\displaystyle \frac{{180}^{\circ }-{36}^{\circ }}{2}}={72}^{\circ }⟹\mathrm{\angle }ABF={180}^{\circ }-{72}^{\circ }-2\ast {32}^{\circ }={36}^{\circ }=\mathrm{\angle }FAB⟹DF=AF=BF$

$⟹\mathrm{\angle }FBD={\displaystyle \frac{{180}^{\circ }-({180}^{\circ }-\mathrm{\angle }AFD-\mathrm{\angle }EFB)}{2}}={66}^{\circ }⟹\mathrm{\angle }ABD={66}^{\circ }-{36}^{\circ }={30}^{\circ }$

$⟹\mathrm{△}ABD\sim \mathrm{△}ACB⟹{\displaystyle \frac{BC}{AC}}={\displaystyle \frac{BD}{AB}}⟹AB\cdot BC=AC\cdot BD◼$

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03/20/2022

Let $A_1={1998}^{50}$, $A_{i+1}$ is the sum of all digits of $A_i$ for $i=1,2,\dots$ until $A_n<10$. Find $A_n$.

Solve:

Let $A_i={\displaystyle \sum _{i=0}^k{a}_i\cdot {10}^i}$, then $A_{i+1}={\displaystyle \sum _{i=0}^k{a}_i\equiv A_i\text{ mod 9}}$, and if $A_i\text{ mod 9}=0$, then $A_{i+1}=9$

${1998}^{50}\text{ mod 9}=(1998\text{ mod 9}{)}^{50}\text{ mod 9}=0^{5}0\text{ mod 9}=0$, so $A_n=9$

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03/30/2022

$AB=AC,\mathrm{\angle }ACB=\mathrm{\angle }ABC={40}^{\circ }$ in $\mathrm{△}ABC$. Extend $AB$ to $D$ so that $AD=BC$. Find $\mathrm{\angle }ADC$.

Solve:

Select point $E$ so that $\mathrm{\angle }EAD=\mathrm{\angle }EDA={40}^{\circ }$, connect $AE,CE,DE$.

Since $AD=BC$, $\mathrm{△}ABC\cong \mathrm{△}ADE⟹AC=AB=AE=DE,\mathrm{\angle }AED=\mathrm{\angle }CAB={100}^{\circ }$

$\mathrm{\angle }CAE={100}^{\circ }-{40}^{\circ }={60}^{\circ }⟹\mathrm{△}ACE\text{ is equilateral triangle }⟹CE=DE,\mathrm{\angle }CEA={60}^{\circ }$

$⟹\mathrm{\angle }CDE=\mathrm{\angle }ECD={\displaystyle \frac{{180}^{\circ }-{60}^{\circ }-{100}^{\circ }}{2}}={10}^{\circ }⟹\mathrm{\angle }ADC=\mathrm{\angle }ADE-\mathrm{\angle }CDE={40}^{\circ }-{10}^{\circ }={30}^{\circ }$

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