06/14/2024

$\mathrm{\angle }A={60}^{\circ }$ in $\mathrm{△}ABC$ and $D$ is the Fermat Point of $\mathrm{△}ABC$, i.e., $\mathrm{\angle }AEB=\mathrm{\angle }AEC={120}^{\circ }$. $AD\perp BC$ at $D$. Prove that $AE=2DE$.

Prove:

$\begin{array}{c}\text{Extend }BE\text{ and }CE\text{ to }M,N\text{ respectively such that }AM\perp CM,BN\perp AN\hfill \\ \mathrm{\angle }AEB=\mathrm{\angle }AEC={120}^{\circ }⟹\mathrm{\angle }AEM=\mathrm{\angle }AEN={60}^{\circ }\hfill \\ ⟹ME=NE={\displaystyle \frac{AE}{2}},\mathrm{\angle }MAN={60}^{\circ },\mathrm{\angle }MEN={120}^{\circ }⟹\mathrm{\angle }MAB=\mathrm{\angle }CAN\hfill \\ AN\perp BN,AD\perp BC,AM\perp CM⟹AMDC\text{ is cyclic},ANDB\text{ is cyclic}\hfill \\ \mathrm{\angle }ADN=\mathrm{\angle }ABN={90}^{\circ }-\mathrm{\angle }BAN={90}^{\circ }-({60}^{\circ }-\mathrm{\angle }CAN)={30}^{\circ }+\mathrm{\angle }CAN\hfill \\ \mathrm{\angle }ADM=\mathrm{\angle }ACM={90}^{\circ }-\mathrm{\angle }CAM={90}^{\circ }-({60}^{\circ }+\mathrm{\angle }MAB)={30}^{\circ }-\mathrm{\angle }MAB\hfill \\ ⟹\mathrm{\angle }ADN+\mathrm{\angle }ADM=\mathrm{\angle }MDN={60}^{\circ }={\displaystyle \frac{\mathrm{\angle }MEN}{2}}⟹\hfill \\ E\text{ is the circumcenter of }\mathrm{△}MDN⟹ME=NE=DE={\displaystyle \frac{AE}{2}}◼\hfill \end{array}$

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06/16/2024

$M$ is the midpoint of side $BC$ of $\mathrm{△}ABC$, $D$ is also on side $BC$ such that $\mathrm{\angle }BAD=\mathrm{\angle }CAM$, prove that ${\displaystyle \frac{BD}{CD}}={\displaystyle \frac{A{B}^2}{A{C}^2}}$

Prove: $\begin{array}{c}CM=BM⟹[CAM]=[BAM]\hfill \\ ⟹{\displaystyle \frac{1}{2}}AC\cdot AM\cdot sin\mathrm{\angle }CAM={\displaystyle \frac{1}{2}}AM\cdot AB\cdot sin\mathrm{\angle }BAM\hfill \\ ⟹AC\cdot sin\mathrm{\angle }BAD=AB\cdot sin\mathrm{\angle }CAD\hfill \\ ⟹{\displaystyle \frac{sin\mathrm{\angle }BAD}{sin\mathrm{\angle }CAD}}={\displaystyle \frac{AB}{AC}}⟹{\displaystyle \frac{BD}{CD}}={\displaystyle \frac{[BAD]}{[CAD]}}\hfill \\ ={\displaystyle \frac{{\displaystyle \frac{1}{2}}AB\cdot AD\cdot sin\mathrm{\angle }BAD}{{\displaystyle \frac{1}{2}}AC\cdot AD\cdot sin\mathrm{\angle }CAD}}={\displaystyle \frac{AB\cdot sin\mathrm{\angle }BAD}{AC\cdot sin\mathrm{\angle }CAD}}={\displaystyle \frac{A{B}^2}{A{C}^2}}◼\hfill \end{array}$

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06/17/2024

The incircle of $\mathrm{△}ABC$ touches its sides $BC,CA,AB$ at points $D,E,F$ respectively. The excircles of $\mathrm{△}ABC$ touch the corresponding sides of $\mathrm{△}ABC$ at points $T,U,V$ (these points are inside the segments $BC,CA,AB$). Show that $[DEF]=[TUV]$.

Prove:

$\begin{array}{c}\text{Get the barymetric coordinates of }D,E,F:\hfill \\ \text{Let }BD=BF=x,CD=CE=y,AE=AF=z\hfill \\ ⟹x+y=a,y+z=b,z+x=c\hfill \\ ⟹x={\displaystyle \frac{a-b+c}{2}},y={\displaystyle \frac{a+b-c}{2}},z={\displaystyle \frac{-a+b+c}{2}}⟹\hfill \\ D(0,{\displaystyle \frac{a-b+c}{2}},{\displaystyle \frac{a+b-c}{2}})\hfill \\ E({\displaystyle \frac{-a+b+c}{2}},0,{\displaystyle \frac{a+b-c}{2}})\hfill \\ F({\displaystyle \frac{-a+b+c}{2}},{\displaystyle \frac{a-b+c}{2}},0)\hfill \\ [DEF]=[ABC]\cdot \left|\begin{array}{ccc}0& {\displaystyle \frac{a-b+c}{2}}& {\displaystyle \frac{a+b-c}{2}}\\ {\displaystyle \frac{-a+b+c}{2}}& 0& {\displaystyle \frac{a+b-c}{2}}\\ {\displaystyle \frac{-a+b+c}{2}}& {\displaystyle \frac{a-b+c}{2}}& 0\end{array}\right|\hfill \\ =[ABC]\cdot {\displaystyle \frac{(-a+b+c)(a-b+c)(a+b-c)}{4}}\hfill \end{array}$ $\begin{array}{c}\text{Similarly, get the barycentric coordinates of }T,U,V:\hfill \\ x+y=a,x+c=y+b⟹x={\displaystyle \frac{a+b-c}{2}},y={\displaystyle \frac{a-b+c}{2}}\hfill \\ ⟹T(0,{\displaystyle \frac{a+b-c}{2}},{\displaystyle \frac{a-b+c}{2}})\hfill \\ m+n=b,m+a=n+c⟹m={\displaystyle \frac{-a+b+x}{2}},n={\displaystyle \frac{a+b-c}{2}}\hfill \\ ⟹U({\displaystyle \frac{a+b-c}{2}},0,{\displaystyle \frac{-a+b+c}{2}})\hfill \\ p+q=c,p+b=q+a⟹p={\displaystyle \frac{a-b+c}{2}},q={\displaystyle \frac{-a+b+c}{2}}\hfill \\ ⟹V({\displaystyle \frac{a-b+c}{2}},{\displaystyle \frac{-a+b+c}{2}},0)\hfill \\ ⟹[TUV]=[ABC]\cdot \left|\begin{array}{ccc}0& {\displaystyle \frac{a+b-c}{2}}& {\displaystyle \frac{a-b+c}{2}}\\ {\displaystyle \frac{a+b-c}{2}}& 0& {\displaystyle \frac{-a+b+c}{2}}\\ {\displaystyle \frac{a-b+c}{2}}& {\displaystyle \frac{-a+b+c}{2}}& 0\end{array}\right|\hfill \\ =[ABC]\cdot {\displaystyle \frac{(a+b-c)(a-b+c)(-a+b+c)}{4}}=[DEF]◼\hfill \end{array}$

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06/21/2024

$AE$ is the bisector of $\mathrm{\angle }A$ in right angle $\mathrm{△}ABC$ and $E$ is on side $BC$, $\mathrm{\angle }C={90}^{\circ }$, $D$ is the midpoint of $AE$, $BC=BD,CE=2$, find $BE$.

Solve:

$\begin{array}{c}\text{Make }DF\perp AC\text{ at }F,DG\perp AB\text{ at }G,DH\perp BC\text{ at }H\hfill \\ ⟹DF=DG={\displaystyle \frac{CE}{2}}=1,DH=\sqrt{a^2-1},AC=2\sqrt{a^2-1}\hfill \\ \text{Let }AD=DE=CD=a,BC=BD=x,AB=y\hfill \\ CD=CE,BC=BD⟹\mathrm{△}CDE\sim \mathrm{△}CBD\hfill \\ ⟹{\displaystyle \frac{a}{2}}={\displaystyle \frac{x}{a}}⟹x={\displaystyle \frac{a^2}{2}}\hfill \\ [ABC]=[ACD]+[ABD]+[BCD]=\sqrt{a^2-1}+{\displaystyle \frac{y}{2}}+{\displaystyle \frac{a^2\sqrt{a^2-1}}{4}}\hfill \\ ⟹y=({\displaystyle \frac{a^2}{2}}-2)\sqrt{a^2-1}\hfill \\ A{C}^2+B{C}^2=A{B}^2⟹4(a^2-1)+{\displaystyle \frac{a^4}{4}}=({\displaystyle \frac{a^2}{2}}-2{)}^2(a^2-1)\hfill \\ ⟹a^4-10a^2+8=0⟹a^2=5\pm \sqrt{17}⟹BE={\displaystyle \frac{1+\sqrt{17}}{2}}\hfill \end{array}$ Solve 2 $\begin{array}{c}\text{Let }AD=DE=CD=a,BC=BD=x,\text{ make }DH\perp BC\text{ at }H\hfill \\ \text{easy to see }\mathrm{\angle }BAC=\mathrm{\angle }CDE=\mathrm{\angle }CBD,AE\text{ bisects }\mathrm{\angle }BAC⟹\hfill \\ cos\mathrm{\angle }BAC={\displaystyle \frac{AC}{AB}}={\displaystyle \frac{CE}{BE}}={\displaystyle \frac{2}{x-2}}=cos\mathrm{\angle }CBD={\displaystyle \frac{BH}{BD}}={\displaystyle \frac{x-1}{x}}\hfill \\ ⟹x^2-5x+2=0⟹x={\displaystyle \frac{5\pm \sqrt{17}}{2}}⟹BE={\displaystyle \frac{1+\sqrt{17}}{2}}\hfill \end{array}$

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06/25/2024

A quarter of circle $OAD$ is tangent with line $BC$ and $AB\perp BC,CD\perp BC$, $AB=4,CD=2$, find the area of Circular sector $OAD$

Solve:

$\begin{array}{c}\text{Connect }AE,DE,\text{ let }OF\perp AE\text{ at }F,OG\perp DE\text{ at }G\hfill \\ \text{Let }\mathrm{\angle }AEB=\mathrm{\angle }EOF=\mathrm{\angle }\alpha ,\mathrm{\angle }CED=\mathrm{\angle }EOG=\mathrm{\angle }\beta ⟹\hfill \\ OF=R\cdot cos\alpha ,EF=R\cdot sin\alpha ⟹AE=2R\cdot sin\alpha \hfill \\ ⟹AB=2R\cdot si{n}^2\alpha =4⟹si{n}^2\alpha ={\displaystyle \frac{2}{R}}\hfill \\ OG=R\cdot cos\beta ,EG=R\cdot sin\beta ⟹DE=2R\cdot sin\beta \hfill \\ ⟹CD=2R\cdot si{n}^2\beta =2⟹si{n}^2\beta ={\displaystyle \frac{1}{R}}\hfill \\ AD=\sqrt{2}R⟹FG={\displaystyle \frac{AD}{2}}={\displaystyle \frac{R}{\sqrt{2}}}\hfill \\ \mathrm{\angle }AOD={90}^{\circ }⟹\mathrm{\angle }AED={180}^{\circ }-{\displaystyle \frac{\mathrm{\angle }AOD}{2}}={135}^{\circ }\hfill \\ ⟹cos\mathrm{\angle }AED={\displaystyle \frac{E{F}^2+E{G}^2-F{G}^2}{2\cdot EF\cdot EG}}\hfill \\ ={\displaystyle \frac{R^2\cdot si{n}^2\alpha +R^2\cdot si{n}^2\beta -{\displaystyle \frac{R^2}{2}}}{2R^2\cdot sin\alpha \cdot sin\beta }}=-{\displaystyle \frac{1}{\sqrt{2}}}⟹\hfill \\ sin\alpha \cdot sin\beta ={\displaystyle \frac{R-6}{2\sqrt{2}R}}⟹si{n}^2\alpha \cdot si{n}^2\beta ={\displaystyle \frac{(R-6{)}^2}{8R^2}}={\displaystyle \frac{2}{R^2}}\hfill \\ ⟹(R-6{)}^2=16⟹R=10\text{ or }R=2\text{ (exclude since }si{n}^2\alpha ={\displaystyle \frac{2}{R}})\hfill \\ ⟹[O\stackrel{⌢}{AD}]={\displaystyle \frac{\pi \cdot 100}{4}}=25\pi \hfill \end{array}$ Solve 2:

$\begin{array}{c}\text{Make }DF\perp OE\text{ at }F,AG\perp OE\text{ at }G⟹EF=2,FG=2\hfill \\ \text{Let }OG=x,\text{ easy to see }\mathrm{△}AOG\cong \mathrm{△}DOF⟹\hfill \\ DF=OG=x,AG=OF=x+2,OE=OA=OD=x+4\hfill \\ ⟹O{G}^2+A{G}^2=O{A}^2⟹x^2+(x+2{)}^2=(x+4{)}^2\hfill \\ ⟹x=6⟹OA=10⟹[O\stackrel{⌢}{AD}]={\displaystyle \frac{\pi \cdot 100}{4}}=25\pi \hfill \end{array}$

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