12/03/2021

Two green circles are externally tangent and two purple circles are externally tangent at the same point. And all these four circles are internally tangent to a blue circle. The radiuses of the green circles are 3 and 4. The radius of one purple circle is 2. Find the radius of the other purple circle.

Solve:

Let $A,B$ be the centers of two green circles, $C,D$ the centers of two purple circles, $O$ the tangent point. Let circle $O$ be the unit circle centered at $O$. Get the inversions of all the circles:

Due to the properties of inversion, the green circles pass through the inversion center $O$, so they are converted to line $a,b$, with distances to $O$ as ${\displaystyle \frac{1}{2r_A}},{\displaystyle \frac{1}{2r_B}}$ respectively; similarly, the purple circles are converted to line $c,d$ with distances to $O$ as ${\displaystyle \frac{1}{2r_C}},{\displaystyle \frac{1}{2r_D}}$ respectively. The blue circle does not pass through the inversion center $O$ so it is converted to a circle, still tangent to the mirrors of the four circles, i.e., four lines $a,b,c,d$. So we know that the distance between line $a,b$ is equal to the distance between line $c,d$:

${\displaystyle \frac{1}{2r_A}}+{\displaystyle \frac{1}{2r_B}}={\displaystyle \frac{1}{2r_C}}+{\displaystyle \frac{1}{2r_D}}⟹{\displaystyle \frac{1}{r_A}}+{\displaystyle \frac{1}{r_B}}={\displaystyle \frac{1}{r_C}}+{\displaystyle \frac{1}{r_D}}$

$r_A=3,r_B=4,r_C=2⟹r_D=12$

So the radius of the other purple circle is 12.

[中文] 两个相互外切的绿色圆与两个相互外切的紫色圆的切点在同一点，这四个圆与蓝色圆内切。已知绿色圆半径为3和4，其一紫色圆半径为2，求另一紫色圆半径。

解：

设绿色圆圆心$A,B$和紫色圆圆心$C,D$以及切点$O$如上图所示。做以$O$为圆心的单位圆（下图红色圆），对所有五个圆做关于该单位圆的反演变换：

根据反演变换性质易知，绿圆$A,B$经过反演中心$O$，所以它们关于单位圆$O$的像为直线$a,b$，与反演中心$O$的距离分别为${\displaystyle \frac{1}{2r_A}},{\displaystyle \frac{1}{2r_B}}$，紫圆$C,D$也经过反演中心$O$，所以它们关于单位圆$O$的像为直线$c,d$，与反演中心的距离分别为${\displaystyle \frac{1}{2r_C}},{\displaystyle \frac{1}{2r_D}}$。蓝圆与四圆相切，且不经过反演中心$O$，所以它关于单位圆$O$的像仍然为圆，且与四圆的像相切，即为上图所示虚线蓝圆，可见直线$a,b$的相距与直线$c,d$的相距相等，即

${\displaystyle \frac{1}{2r_A}}+{\displaystyle \frac{1}{2r_B}}={\displaystyle \frac{1}{2r_C}}+{\displaystyle \frac{1}{2r_D}}⟹{\displaystyle \frac{1}{r_A}}+{\displaystyle \frac{1}{r_B}}={\displaystyle \frac{1}{r_C}}+{\displaystyle \frac{1}{r_D}}$

$r_A=3,r_B=4,r_C=2⟹r_D=12$

故所求紫色圆半径为12。

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12/04/2021

Rectangle $ABCD$ has two points $E,F$ on side $BC,CD$, and $\mathrm{△}AEF$ is equilateral. Prove $S_{\mathrm{△}CEF}=S_{\mathrm{△}ADF}+S_{\mathrm{△}ABE}$

Prove:

Let $S_{\mathrm{△}AEF}=S_0,S_{\mathrm{△}ADF}=S_1,S_{\mathrm{△}ABE}=S_2,S_{\mathrm{△}CEF}=S_3$. So we need prove $S_3=S_1+S_2$.

Extend $AB$ to $A^{\prime }$ so that $AB=B{A}^{\prime }$, extend $DC$ to $D^{\prime }$ so that $DC=C{D}^{\prime }$. Extended $AE$ and $A^{\prime }{D}^{\prime }$ intersect at $G$. Connect $FG$.

Easy to see that $AB=B{A}^{\prime }⟹AE=EG⟹S_{\mathrm{△}AEF}=S_{\mathrm{△}FEG}=S_0,S_{EB{A}^{\prime }G}=3S_{\mathrm{△}ABE}=3S_2$

$\mathrm{\angle }FAE={60}^{\circ },AE=EF=AF⟹\mathrm{\angle }EFG=\mathrm{\angle }AGF={30}^{\circ }⟹\mathrm{\angle }AFG={90}^{\circ },GF=\sqrt{3}AF$

$⟹\mathrm{\angle }DAF=\mathrm{\angle }{D}^{\prime }FG⟹S_{\mathrm{△}{D}^{\prime }FG}=3S_{\mathrm{△}ADF}=3S_1$

$⟹S_{A{A}^{\prime }{D}^{\prime }D}=2S_{ABCD}=2(S_0+S_1+S_2+S_3)=2S_0+S_1+3S_1+4S_2⟹S_3=S_1+S_2◼$

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12/05/2021

Given any five integers, prove that there must be four of them as $a,b,c,d$ so that 28 divides $(a^2-b^2)(c^2-d^2)$

Proof: All integers mod by $7$ we get $0,1,2,3,4,5,6$, so their squares mod by $7$ will be $0,1,4,9\equiv 2,16\equiv 2,25\equiv 4,36\equiv 1=0,1,2,4$.

So pick any $5$ from them will at least have two integers that the difference of their squares is $0$ mod 7. Take them as $a,b$.

All integers mod by $4$ we get $0,1,2,3$, so their squares mod by $4$ will be $0,1,4\equiv 0,9\equiv 1=0,1$. For the rest of the three integers after above step, there must be two of them so that the difference of their squares is $0$ mod 4. Take them as $c,d$.

So there will be $(a^2-b^2)(c^2-d^2)\equiv 0$ mod $28$. $◼$

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12/06/2021

In $\mathrm{△}ACD,\mathrm{\angle }A={84}^{\circ },\mathrm{\angle }ACD={42}^{\circ }$. Now extend $AD$ to $B$ so that $AC=BD$. Find $\mathrm{\angle }BCD$.

Solve 1:

Make $A^{\prime }$ on $AD$ so that $\mathrm{\angle }AC{A}^{\prime }={12}^{\circ }⟹AC=A^{\prime }C,\mathrm{\angle }{A}^{\prime }AC={84}^{\circ },\mathrm{\angle }ACD={30}^{\circ }$.

Let $E$ be the circumcenter of $\mathrm{△}{A}^{\prime }CD⟹A^{\prime }E=CE=DE,\mathrm{\angle }CED=2\ast \mathrm{\angle }{A}^{\prime }AC={168}^{\circ }⟹\mathrm{\angle }EDC=\mathrm{\angle }ECD=6^{\circ }$

$⟹\mathrm{\angle }ED{A}^{\prime }={54}^{\circ }+6^{\circ }={60}^{\circ }⟹\mathrm{△}{A}^{\prime }DE$ is equilateral triangle.

Make equilateral $\mathrm{△}{A}^{\prime }CF$, so that $A^{\prime }C=CF=A^{\prime }F=BD$, $A^{\prime }F$ and $CG$ intersect at $G$. Easy to see that $CG\perp A^{\prime }F⟹DF=A^{\prime }D,\mathrm{\angle }DF{A}^{\prime }=\mathrm{\angle }D{A}^{\prime }F=\mathrm{\angle }E{A}^{\prime }C=\mathrm{\angle }EC{A}^{\prime }={36}^{\circ }$

Now make $H$ on $A^{\prime }C$ so that $CE=CH=DE=A^{\prime }E⟹\mathrm{\angle }CHE=\mathrm{\angle }CEH={72}^{\circ }⟹\mathrm{\angle }E{A}^{\prime }H=\mathrm{\angle }{A}^{\prime }EH={36}^{\circ }⟹A^{\prime }H=EH$

$⟹\mathrm{△}{A}^{\prime }DF\sim \mathrm{△}{A}^{\prime }EC⟹{\displaystyle \frac{A^{\prime }D}{AF}}={\displaystyle \frac{A^{\prime }H}{A^{\prime }E}}⟹{\displaystyle \frac{A^{\prime }D+AF}{AF}}={\displaystyle \frac{A^{\prime }H+A^{\prime }E}{A^{\prime }E}}⟹{\displaystyle \frac{A^{\prime }D+BD}{AC}}={\displaystyle \frac{A^{\prime }H+CH}{A^{\prime }D}}$

$⟹{\displaystyle \frac{A^{\prime }B}{A^{\prime }C}}={\displaystyle \frac{A^{\prime }C}{A^{\prime }D}}⟹\mathrm{△}{A}^{\prime }BC\sim \mathrm{△}{A}^{\prime }CD⟹\mathrm{\angle }{A}^{\prime }BC=\mathrm{\angle }{A}^{\prime }CD={30}^{\circ }⟹\mathrm{\angle }BCD={54}^{\circ }-{30}^{\circ }={24}^{\circ }$

Solve 2:

Make $A^{\prime }$ on $AD$ so that $\mathrm{\angle }AC{A}^{\prime }={12}^{\circ }⟹AC=A^{\prime }C,\mathrm{\angle }{A}^{\prime }AC={84}^{\circ },\mathrm{\angle }{A}^{\prime }CD={30}^{\circ }$.

Make equilateral $\mathrm{△}{A}^{\prime }CF$, so that $A^{\prime }C=CF=A^{\prime }F=BD$, $A^{\prime }F$ and $CG$ intersect at $G$. Easy to see that $CG\perp A^{\prime }F⟹DF=A^{\prime }D,\mathrm{\angle }DF{A}^{\prime }=\mathrm{\angle }D{A}^{\prime }F={36}^{\circ }⟹\mathrm{\angle }BDF={72}^{\circ }$

Make $D^{\prime }$ on $BD$ so that $F{D}^{\prime }=FD⟹\mathrm{\angle }D{D}^{\prime }F={72}^{\circ }⟹\mathrm{\angle }DF{D}^{\prime }={36}^{\circ }⟹\mathrm{\angle }A{D}^{\prime }F=\mathrm{\angle }{A}^{\prime }F{D}^{\prime }={72}^{\circ }$

$⟹A^{\prime }F=A^{\prime }{D}^{\prime }=AC=BD⟹B{D}^{\prime }=A^{\prime }D=FD=F{D}^{\prime }⟹\mathrm{\angle }{D}^{\prime }BF={36}^{\circ }⟹\mathrm{\angle }BFD=\mathrm{\angle }BDF={72}^{\circ }$

$⟹BD=BF=AC=A^{\prime }C=CF⟹\mathrm{\angle }BFC=\mathrm{\angle }BFD+\mathrm{\angle }DF{A}^{\prime }+\mathrm{\angle }{A}^{\prime }FC={168}^{\circ }⟹\mathrm{\angle }BCF=6^{\circ }$

$⟹\mathrm{\angle }BCD={60}^{\circ }-{30}^{\circ }-6^{\circ }={24}^{\circ }$

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12/12/2021

Let $a$, $b$ and $n$ be positive integers such that $n\ge 2$ and $(ab{)}^{n-1}+1$ divides $a^n+b^n$. Show that ${\displaystyle \frac{a^n+b^n}{a{b}^{n-1}+1}}$ is a perfect $n^{th}$ power of an integer. i.e., $k={\displaystyle \frac{a^n+b^n}{(ab{)}^{n-1}+1}}\in \mathbb{N}⟹\mathrm{\exists }c\in \mathbb{N},k=c^n$. (special case when $n=2$: 1988 IMO Problem 6)

Prove:

(1) When $a=b$:

$k={\displaystyle \frac{2a^n}{a^{2n-2}+1}}\in \mathbb{N}⟹2a^n=k{a}^{2n-2}+k⟹2=k{a}^{n-2}+{\displaystyle \frac{k}{a^n}}$

$n\ge 2,n\in \mathbb{N},a\in \mathbb{N},k\in \mathbb{N}⟹n-2\ge 1⟹a^{n-2}\ge 1⟹k{a}^{n-2}\ge 1$

$⟹k{a}^{n-2}=1,{\displaystyle \frac{k}{a^n}}=1⟹a^{2n-2}=1⟹a=b=1,k=1$

(2) When $a\ne b$, WLOG, suppose $a>b$

(2.1) When $n=2$

Suppose $k$ is a non-square positive integer and there exist positive integers $(a,b)$ so that $k={\displaystyle \frac{a^2+b^2}{ab+1}}$. Among these positive integers, suppose $(A,B)$ are the ones satisfying $k={\displaystyle \frac{A^2+B^2}{AB+1}}$ and such that $A+B$ is minimized, and WLOG, assume $A>B$.

Fixing $B$, then $A$ is one root $x_1=A$ of equation $x^2-(kB)x+(B^2-k)=0$. Then we know the other root $x_2$ satisfies $x_2=kB-A$ and $x_2={\displaystyle \frac{B^2-k}{A}}$.

$x_2=kB-A⟹x_2\in \mathbb{Z}$. $k$ is not a perfect square and $x_2={\displaystyle \frac{B^2-k}{A}}⟹x_2\ne 0$

${\displaystyle \frac{x_2^2+B^2}{x_{2}B+1}}=k>0⟹x_2>0⟹x_2\in \mathbb{N}$

$A>B⟹A^2>B^2>B^2-k⟹A>{\displaystyle \frac{B^2-k}{A}}$

$⟹x_2={\displaystyle \frac{B^2-k}{A}}<x_1=A⟹x_2+B<A+B$

This means, with $B$ fixed, we have $x_2$ as a smaller root than $x_1=A$ and we get even smaller $x_2+B<A+B$. This contradicts the minimality of $A+B$.

This means $x_2$ can be smaller and smaller until it reaches $0$ and that means $k=B^2$. The original assumption is not correct and $k$ is a perfect square.

(2.2) When $n>2$

We have $n>2,a>b,k={\displaystyle \frac{a^n+b^n}{(ab{)}^{n-1}+1}},a,b,k,n\in \mathbb{N}⟹b^n-k=(k{b}^{n-1}-a)a^{n-1}$

(2.2.1) If $k>b^n$, then $k>k-b^n=a^{n-1}(a-k{b}^{n-1})>0⟹a-k{b}^{n-1}>0⟹a-k{b}^{n-1}\ge 1$

$⟹k\ge a^{n-1}⟹a>k{b}^{n-1}\ge k\ge a^{n-1}⟹a>a^{n-1}$ this is not possible

(2.2.2) If $k<b^n$, and since $b<a$ then $k{b}^{n-1}-a={\displaystyle \frac{b^n-k}{a^{n-1}}}<{\displaystyle \frac{b^n}{a^{n-1}}}<{\displaystyle \frac{b^n}{b^{n-1}}}=b$

$⟹k{b}^{n-1}<a+b⟹(k-1)b^{n-1}<a+b-b^{n-1}\le a⟹(k-1)b^{n-1}<a$

and $k<b^n⟹b^n-k=(k{b}^{n-1}-a)a^{n-1}>0⟹k{b}^{n-1}-a>0⟹k{b}^{n-1}-a\ge 1$

if $k>1$, then $2\le k⟹1\le k-1⟹b^{n-1}\le (k-1)b^{n-1}<a⟹b^{n-1}<a$

$⟹b^n-k=(k{b}^{n-1}-a)a^{n-1}\ge a^{n-1}>(b^{n-1}{)}^{n-1}\ge b^n⟹b^n-k>b^n$ this is not possible, so the only possible situation is $k=1=1^n$ , then the statement is proved true for this case.

(2.2.3) If $k=b^n$ then the statement is proved true for this case as well.

References:

• The case $n=2$ of this problem is mentioned in Vieta Jumping Wiki
• The case $n=2$ of this problem is mentioned in book Problem-Solving Strategies by Arthur Engel, in Chapter 6, E15.

• The case $n=2$ of this problem is also mentioned in book Modern Olympiad Number Theory by Aditya Khurmi, in Chapter 4, Example 4.7.1.

• Alternative proof that $(a^2+b^2)/(ab+1)$ is a square when it’s an integer
• When is $f(a,b)={\displaystyle \frac{a^2+b^2}{1+ab}}$ a perfect square rational number?
• AOPS link 1
• AOPS link 2
• AOPS link 3
• What is the algebraic intuition behind Vieta jumping in IMO1988 Problem 6?
• Underlying structure behind the infamous IMO 1988 Problem 6
• When do Pell equation results imply applicability of the “Vieta jumping”-method to a given conic?
• The Legend of Question Six - Numberphile

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12/18/2021

$AB=AC,\mathrm{\angle }CAB={90}^{\circ }\text{ in }\mathrm{△}ABC$, $D$ is inside $\mathrm{△}ABC$ and $E$ is the midpoint of $CD$ so that $\mathrm{\angle }CAE=\mathrm{\angle }ABD$. Show that $\mathrm{\angle }DAE={45}^{\circ }$.

Prove:

Extend $BD$ so it and $AE$ intersect at $F$. Extend $AE$ to $G$ so that $CG\perp AG$.

$\mathrm{\angle }CAE+\mathrm{\angle }EAB={90}^{\circ },\mathrm{\angle }CAE=\mathrm{\angle }ABD⟹\mathrm{\angle }ABD+\mathrm{\angle }EAB={90}^{\circ }⟹AF\perp BF$

$AC=AB,\mathrm{\angle }CAE=\mathrm{\angle }ABD,\mathrm{\angle }AGC=\mathrm{\angle }AFB={90}^{\circ }⟹\mathrm{△}ACG\cong \mathrm{△}BAF⟹CG=AF$

$CE=DE,\mathrm{\angle }DFE=\mathrm{\angle }CGE={90}^{\circ },\mathrm{\angle }CEG=\mathrm{\angle }DEF⟹\mathrm{△}DEF\cong \mathrm{△}CEG⟹CG=DF$

$⟹AF=DF⟹\mathrm{\angle }DAE={45}^{\circ }◼$

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12/20/2021

Find $\mathrm{\angle }BAC$ in following figure:

Solve:

Connect $EF$:

Easy to see that $AE\perp BD⟹\mathrm{△}DEG\cong \mathrm{△}BEG$

$⟹\mathrm{\angle }EBG=\mathrm{\angle }EDG={30}^{\circ },DE=BE,DG=BG⟹\mathrm{△}ADG\cong \mathrm{△}ABG$

$⟹\mathrm{\angle }BAE=\mathrm{\angle }DAE={40}^{\circ }=\mathrm{\angle }EDF⟹ADEF\text{ are cyclic }⟹\mathrm{\angle }AEF=\mathrm{\angle }ADF={40}^{\circ }$

$⟹AF=EF,\mathrm{\angle }BEF=\mathrm{\angle }FCE={20}^{\circ }⟹CF=EF,\mathrm{\angle }EFB={180}^{\circ }-{20}^{\circ }-{50}^{\circ }-{30}^{\circ }={80}^{\circ }$

$⟹BE=EF=DE=AF=CF⟹AF=CF⟹\mathrm{\angle }CAB={\displaystyle \frac{\mathrm{\angle }CFB}{2}}={\displaystyle \frac{{60}^{\circ }}{2}}={30}^{\circ }$

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12/28/2021

In quadrilateral $ABCD$, $AD=2CD,BC\perp CD,\mathrm{\angle }ADC=2\mathrm{\angle }A$. Prove: $AB=BD$

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12/29/2021

Solve:

Let $E$ be the circumcenter of $\mathrm{△}BCD$, easy to see that

$BE=DE=CE,\mathrm{\angle }BEC=2\ast \mathrm{\angle }BCD={60}^{\circ }⟹BD=DE,\mathrm{\angle }BDE={60}^{\circ }$

$⟹\mathrm{\angle }ADE={360}^{\circ }-{150}^{\circ }-{60}^{\circ }={150}^{\circ }=\mathrm{\angle }ADB⟹\mathrm{△}ABD\cong \mathrm{△}AED$

$⟹\mathrm{\angle }BAD=\mathrm{\angle }EAD,AB=AE=AC⟹\mathrm{△}ABE\cong \mathrm{△}ACE$

$⟹\mathrm{\angle }BAE=\mathrm{\angle }CAE⟹\mathrm{\angle }BAD={\displaystyle \frac{\mathrm{\angle }CAD}{3}}={\displaystyle \frac{{39}^{\circ }}{3}}={13}^{\circ }$

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