01/02/2024

$AB$ and $CD$ are two tangent lines of $\odot O$. $AB=8,CD=5$, $AD\cap BC=E$, and $E$ is on $\odot O$. Find $AC$.

Solve 1:

$\begin{array}{c}\text{WLOG, suppose }AB>CD⟹C\text{ is inside the circumcircle of }\mathrm{△}ABD\hfill \\ \text{Extend }BC,DC\text{ and intersect the circumcircle of }\mathrm{△}ABD\text{ at }H,G\text{ respectively}\hfill \\ \text{Let }AB=m,\mathrm{\angle }ABH=\mathrm{\angle }ADB=\mathrm{\angle }AHB⟹AB=AH=m\hfill \\ \mathrm{\angle }ADG=\mathrm{\angle }DBH⟹\mathrm{\angle }ABD=\mathrm{\angle }GDB⟹\mathrm{\angle }AGD+\mathrm{\angle }GDB={180}^{\circ }\hfill \\ ⟹AG\parallel BD⟹ABDG\text{ is an isosceles trapezoid}⟹AB=GD=m\hfill \\ \text{Let }CD=n⟹CG=m-n，\text{applying Intersecting chords theorem:}\hfill \\ ⟹CH\cdot BC=CD\cdot GC⟹CH\cdot BC=n(m-n)\hfill \\ \text{Applying Stewart's Theorem in }\mathrm{△}ABH⟹A{C}^2={\displaystyle \frac{A{H}^{2}BC+A{B}^{2}CH}{CH+BC}}-CH\cdot BC\hfill \\ ⟹A{C}^2={\displaystyle \frac{m^2\cdot BH}{BH}}-CH\cdot BC=m^2-n(m-n)⟹AC=\sqrt{m^2-mn+n^2}\hfill \\ \text{When }m=8,n=5,AC=\sqrt{8^2+5^5-40}=7\hfill \end{array}$

Solve 2:

$\begin{array}{c}\text{Let }F\text{ be a point in the circle such that }CF=CD,AF=AB\hfill \\ \text{Let }\mathrm{\angle }ABC=\mathrm{\angle }\alpha =\mathrm{\angle }BDE,\mathrm{\angle }ADC=\mathrm{\angle }\beta =\mathrm{\angle }DBE\hfill \\ \text{Let }\mathrm{\angle }CFE=\mathrm{\angle }\delta ,CD=CF=CE\cdot CB⟹\mathrm{\angle }AFB=\mathrm{\angle }\alpha +\mathrm{\angle }\delta \hfill \\ ⟹\mathrm{\angle }EBF=\mathrm{\angle }\delta ⟹\mathrm{\angle }DBF=\mathrm{\angle }\beta -\mathrm{\angle }\delta \hfill \\ \text{Let }\mathrm{\angle }AFE=\mathrm{\angle }\gamma ,AB=AF=AE\cdot AD⟹\mathrm{\angle }CFD=\mathrm{\angle }\beta +\mathrm{\angle }\gamma \hfill \\ ⟹\mathrm{\angle }EDF=\mathrm{\angle }\gamma ⟹\mathrm{\angle }BDF=\mathrm{\angle }\alpha -\mathrm{\angle }\gamma \hfill \\ ⟹\mathrm{\angle }BD={180}^{\circ }-(\mathrm{\angle }\beta -\mathrm{\angle }\delta )-(\mathrm{\angle }\alpha -\mathrm{\angle }4)\hfill \\ ⟹{360}^{\circ }=\mathrm{\angle }\delta +\mathrm{\angle }\gamma +(\mathrm{\angle }\alpha +\mathrm{\angle }\delta )+(\mathrm{\angle }\beta +\mathrm{\angle }\gamma )+{180}^{\circ }-(\mathrm{\angle }\beta -\mathrm{\angle }\delta )-(\mathrm{\angle }\alpha -\mathrm{\angle }\gamma )\hfill \\ ={180}^{\circ }+3(\mathrm{\angle }\delta +\mathrm{\angle }\gamma )⟹\mathrm{\angle }\delta +\mathrm{\angle }\gamma =\mathrm{\angle }AFC={60}^{\circ }\hfill \\ ⟹cos(\mathrm{\angle }AFC)={\displaystyle \frac{A{F}^2+C{F}^2-A{C}^2}{2AF\cdot CF}}={\displaystyle \frac{1}{2}}⟹A{C}^2=A{B}^2+C{D}^2-AB\cdot CD\hfill \end{array}$

Note:

• Follow above conclusion, $F$ can be proven to be the first Isodynamic Point of $\mathrm{△}BDE$ since $\mathrm{\angle }DFE=\mathrm{\angle }DBE+{60}^{\circ }$, $\mathrm{\angle }BFE=\mathrm{\angle }BDE+{60}^{\circ }$
• Properties of isodynamic point of a triangle:
  • The first isodynamic point is the isogonal conjugate of the Fermat point of the triangle (the first isogonic center)
  • The second isodynamic point is the isogonal conjugate of the second Fermat point of the triangle (the second isogonic center)

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01/11/2024

$O$ is the intersection of $AD,BE$ and $CF$ in hexagon $ABCDEF$, and $S_{\mathrm{△}ABO}=12m^2$, $S_{\mathrm{△}BCO}=6m^2$, $S_{\mathrm{△}CDO}=8m^2$, $S_{\mathrm{△}DEO}=4m^2$, $S_{\mathrm{△}EFO}=2m^2$, find $S_{\mathrm{△}FAO}$

$\begin{array}{c}{S}_{\mathrm{△}ABO}={\displaystyle \frac{1}{2}}AO\cdot BO\cdot sin\mathrm{\angle }AOB,S_{\mathrm{△}DEO}={\displaystyle \frac{1}{2}}DO\cdot EO\cdot sin\mathrm{\angle }DOE={\displaystyle \frac{1}{2}}DO\cdot EO\cdot sin\mathrm{\angle }AOB\hfill \\ S_{\mathrm{△}BCO}={\displaystyle \frac{1}{2}}BO\cdot CO\cdot sin\mathrm{\angle }BOC,S_{\mathrm{△}EFO}={\displaystyle \frac{1}{2}}EO\cdot FO\cdot sin\mathrm{\angle }EOF={\displaystyle \frac{1}{2}}EO\cdot FO\cdot sin\mathrm{\angle }BOC\hfill \\ S_{\mathrm{△}CDO}={\displaystyle \frac{1}{2}}CO\cdot DO\cdot sin\mathrm{\angle }COD,S_{\mathrm{△}FAO}={\displaystyle \frac{1}{2}}FO\cdot AO\cdot sin\mathrm{\angle }FOA={\displaystyle \frac{1}{2}}FO\cdot AO\cdot sin\mathrm{\angle }COD\hfill \\ ⟹S_{\mathrm{△}ABO}\cdot S_{\mathrm{△}CDO}\cdot S_{\mathrm{△}EFO}=S_{\mathrm{△}BCO}\cdot S_{\mathrm{△}DEO}\cdot S_{\mathrm{△}FAO}\hfill \\ ⟹S_{\mathrm{△}FAO}={\displaystyle \frac{S_{\mathrm{△}ABO}\cdot S_{\mathrm{△}CDO}\cdot S_{\mathrm{△}EFO}}{S_{\mathrm{△}BCO}\cdot S_{\mathrm{△}DEO}}}={\displaystyle \frac{12\ast 8\ast 2}{6\ast 4}}=8m^2\hfill \end{array}$

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01/28/2024

Inscribed $\odot O_1$ is tangent to semi-circle $\odot O$ and its diameter $AB$. Find $tan(\mathrm{\angle }{O}_{1}AB)+tan(\mathrm{\angle }{O}_{1}BA)$.

Solve:

$\begin{array}{c}\text{Let }\mathrm{\angle }{O}_{1}AB=\alpha ,\mathrm{\angle }{O}_{1}BA=\beta ,AO=R,C{O}_1=r,OC=t\hfill \\ ⟹O{O}_1^2=O{C}^2+O_1{C}^2⟹(R-r{)}^2=r^2+t^2\hfill \\ ⟹R^2-t^2=2Rr\hfill \\ tan(\mathrm{\angle }{O}_{1}AB)+tan(\mathrm{\angle }{O}_{1}BA)={\displaystyle \frac{r}{R+t}}+{\displaystyle \frac{r}{R-t}}\hfill \\ ={\displaystyle \frac{2Rr}{R^2-r^2}}=1\hfill \end{array}$

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01/29/2024

$A_1{B}_1,A_2{B}_2$ are focal chords of parabola. $D=A_1{A}_2\cap B_1{B}_2$. Prove that $D$ is on the directrix of the parabola.

Prove:

$\begin{array}{c}\text{Suppose }{A}_1^{\prime },A_2^{\prime }\text{ are the projections of }{A}_1,A_2\text{ on the directrix}\hfill \\ D\text{ is the intersection between extended }{A}_1{A}_2\text{ and the directrix}.\hfill \\ \mathrm{△}D{A}_1{A}_1^{\prime }\sim \mathrm{△}D{A}_2{A}_2^{\prime }⟹{\displaystyle \frac{D{A}_1}{D{A}_2}}={\displaystyle \frac{A_1{A}_1^{\prime }}{A_2{A}_2^{\prime }}}={\displaystyle \frac{F{A}_1}{F{A}_2}}\hfill \\ \text{Via the exterior angle bisector theorem}⟹\mathrm{\angle }{A}_{2}FD=\mathrm{\angle }DF{B}_1\hfill \\ ⟹D\text{ is on bisector of }\mathrm{△}{A}_{2}F{B}_1\hfill \\ \text{Similarly, let}{D}^{\prime }\text{ be the intersection between extended }{B}_1{B}_2\text{ and the directrix}\hfill \\ ⟹D^{\prime }\text{ is on disector of }\mathrm{△}{A}_{2}F{B}_1⟹D=D^{\prime }⟹D=A_1{A}_2\cap B_1{B}_2◼\hfill \end{array}$

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01/30/2024

$F$ is the focus of a parabola, $A,B$ are two points on the parabola and $S$ is the intersection between the two tangent lines of the parabola at $A,B$. $A^{\prime },B^{\prime }$ are the projections of $A,B$ on the directrix. Prove that:

(1) $S$ is the circumcenter of $\mathrm{△}F{A}^{\prime }{B}^{\prime }$

(2) $\mathrm{\angle }FBS=\mathrm{\angle }F{B}^{\prime }{A}^{\prime }=\mathrm{\angle }FSA,\mathrm{\angle }FAS=\mathrm{\angle }F{A}^{\prime }{B}^{\prime }=\mathrm{\angle }FSB$

(3) $F$ is on line $AB$ if and only if $S$ is on line $A^{\prime }{B}^{\prime }$ and in this case, $SA\perp SB$

Prove:

$\begin{array}{c}\boxed{{\displaystyle \text{Reflective Properties of parabola:}}}\hfill \\ \text{Let }A\text{ be a point on a parabola}\hfill \\ A^{\prime }\text{ be the projection of }A\text{ on the directrix}\hfill \\ F\text{ be the focus of the parabola}\hfill \\ ⟹\text{the tangent line of the parabola at }A\text{ bisects }\mathrm{\angle }FA{A}^{\prime }\hfill \\ \text{Prove: }\text{suppose the bisector of }\mathrm{\angle }FA{A}^{\prime }\text{ intersects the parabola in another point }B\hfill \\ \text{Let }{B}^{\prime }\text{ be the projection of }B\text{ on the directrix}⟹BF=B{B}^{\prime }\hfill \\ B\text{ is on bisector of }\mathrm{\angle }FA{A}^{\prime }⟹BF=B{A}^{\prime }⟹B{B}^{\prime }=B{A}^{\prime }\text{ which is impossible}\hfill \\ ⟹A\text{ is the only point of the parabola that is on the bisector of }\mathrm{\angle }FA{A}^{\prime }\hfill \\ ⟹\text{the bisector of }\mathrm{\angle }FA{A}^{\prime }\text{ is the tangent line of parabola at }A◼\hfill \\ ⟹SF=S{A}^{\prime }=S{B}^{\prime }⟹S\text{ is the circumcenter of }\mathrm{△}S{A}^{\prime }{B}^{\prime }◼\hfill \\ ⟹\mathrm{\angle }FS{B}^{\prime }=2\mathrm{\angle }F{A}^{\prime }{B}^{\prime }=2\mathrm{\angle }FSB⟹\mathrm{\angle }F{A}^{\prime }{B}^{\prime }=\mathrm{\angle }FSB\hfill \\ ⟹\mathrm{\angle }FAS=\mathrm{\angle }{A}^{\prime }AS={90}^{\circ }-\mathrm{\angle }A{A}^{\prime }F=\mathrm{\angle }F{A}^{\prime }{B}^{\prime }=\mathrm{\angle }FSB◼\hfill \\ \text{Similarly: }\mathrm{\angle }FBS=\mathrm{\angle }F{B}^{\prime }{A}^{\prime }=\mathrm{\angle }FSA◼\hfill \\ \text{Let }\mathrm{\angle }FSB=\mathrm{\angle }BS{B}^{\prime }=\mathrm{\angle }F{A}^{\prime }{B}^{\prime }=\mathrm{\angle }FAS=\mathrm{\angle }{A}^{\prime }AS=\alpha \hfill \\ \mathrm{\angle }FSA=\mathrm{\angle }AS{A}^{\prime }=\mathrm{\angle }F{B}^{\prime }{A}^{\prime }=\mathrm{\angle }FBS=\mathrm{\angle }{B}^{\prime }BS=\beta \hfill \\ \mathrm{\angle }AFB=\mathrm{\angle }AFS+\mathrm{\angle }BFS={180}^{\circ }-\alpha -\beta +{180}^{\circ }-{180}^{\circ }-\alpha -\beta =\mathrm{\angle }{A}^{\prime }S{B}^{\prime }\hfill \\ ⟹\mathrm{\angle }AFB={180}^{\circ }\text{ if and only if }\mathrm{\angle }{A}^{\prime }S{B}^{\prime }={180}^{\circ }\hfill \\ ⟹F\text{ is on }AB\text{ if and only if }S\text{ is on }{A}^{\prime }{B}^{\prime }◼\hfill \\ S\text{ is on }{A}^{\prime }{B}^{\prime }⟹\mathrm{\angle }{A}^{\prime }S{B}^{\prime }={180}^{\circ }\text{ if and only if }\alpha +\beta ={90}^{\circ }⟹SA\perp SB◼\hfill \end{array}$

Notes:

• Two Tangents to Parabola on cut-the-knot site
• The previous problem on StackExchange
• Optical Property of the parabola from the book “Geometry of Conics” Page 15

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