2022/06/07

$a,b,c>0,a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}=1$, prove:

(1) $abc\le {\displaystyle \frac{1}{9}}$ (2) ${\displaystyle \frac{a}{b+c}}+{\displaystyle \frac{b}{c+a}}+{\displaystyle \frac{c}{a+b}}\le {\displaystyle \frac{1}{2\sqrt{abc}}}$

prove: $\begin{array}{c}\text{(1)}\hfill \\ G_3=\sqrt[3]{x_1{x}_2{x}_3},A_3={\displaystyle \frac{x_1+x_2+x_3}{3}},Gn\le An\hfill \\ a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}=1⟹{\displaystyle \frac{a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}}{3}}={\displaystyle \frac{1}{3}}\hfill \\ ⟹\sqrt[3]{a^{\frac{3}{2}}{b}^{\frac{3}{2}}{c}^{\frac{3}{2}}}=(abc{)}^{\frac{1}{2}}\le {\displaystyle \frac{1}{3}}\hfill \\ ⟹abc\le {\displaystyle \frac{1}{9}}◼\hfill \\ \text{(2)}\hfill \\ {\displaystyle \frac{a}{b+c}}+{\displaystyle \frac{b}{c+a}}+{\displaystyle \frac{c}{a+b}}\le {\displaystyle \frac{a}{2\sqrt{bc}}}+{\displaystyle \frac{b}{2\sqrt{ca}}}+{\displaystyle \frac{c}{2\sqrt{ab}}}\hfill \\ ⟹{\displaystyle \frac{a}{b+c}}+{\displaystyle \frac{b}{c+a}}+{\displaystyle \frac{c}{a+b}}\le {\displaystyle \frac{a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}}{2\sqrt{abc}}}={\displaystyle \frac{1}{2\sqrt{abc}}}◼\hfill \end{array}$

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2022/06/21

In $\mathrm{△}ABC$, $D$ is a point on $BC$ so that $\mathrm{\angle }DAC={60}^{\circ },\mathrm{\angle }ACB={40}^{\circ },AC=BD$, find $\mathrm{\angle }ABC$.

Solve:

$\begin{array}{c}\text{Easy to see }\mathrm{\angle }ADC={180}^{\circ }-{60}^{\circ }-{40}^{\circ }={80}^{\circ }\hfill \\ \text{Extend }AD\text{ to }E\text{ so that }\mathrm{△}ACE\text{ is equilateral}.\hfill \\ ⟹AE=AC=CE,\mathrm{\angle }DCE={20}^{\circ }\hfill \\ \text{Make }F\text{ on }BC\text{ so that }CE=EF\hfill \\ ⟹CE=CF=AC=BD,\mathrm{\angle }CEF=\mathrm{\angle }CFE={80}^{\circ }=\mathrm{\angle }ADC=\mathrm{\angle }FDE\hfill \\ ⟹BF+DF=CD+DF,\mathrm{\angle }FED={20}^{\circ },EF=DE,\mathrm{\angle }BFE=\mathrm{\angle }CDE\hfill \\ ⟹BF=CD\hfill \\ ⟹\mathrm{△}BEF\cong \mathrm{△}CED\hfill \\ ⟹BE=CE=AE\hfill \\ ⟹E\text{ is the circumcenter of }\mathrm{△}ABC\hfill \\ ⟹\mathrm{\angle }ABC={\displaystyle \frac{\mathrm{\angle }AEC}{2}}={30}^{\circ }\hfill \end{array}$

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2022/06/24

In quadrileteral $ABCD$, $BC=CD,\mathrm{\angle }ABD={30}^{\circ },\mathrm{\angle }ACD={38}^{\circ },\mathrm{\angle }BAC={22}^{\circ }$, find $\mathrm{\angle }CAD$.

Solve:

$\begin{array}{c}\mathrm{\angle }CDB=\mathrm{\angle }BAC+\mathrm{\angle }ABD-\mathrm{\angle }ACD={14}^{\circ }=\mathrm{\angle }CBD\hfill \\ ⟹\mathrm{\angle }BCD={180}^{\circ }-2\ast {14}^{\circ }-{38}^{\circ }={114}^{\circ }\hfill \\ \text{Make }F\text{ on }AB\text{ so that }CF=BC=CD\hfill \\ ⟹\mathrm{\angle }CFB=\mathrm{\angle }CBF={30}^{\circ }+{14}^{\circ }={44}^{\circ }\hfill \\ ⟹\mathrm{\angle }FCB={180}^{\circ }-2\ast {44}^{\circ }={92}^{\circ }\hfill \\ ⟹\mathrm{\angle }ACF={114}^{\circ }-{92}^{\circ }={22}^{\circ }=\mathrm{\angle }BAC\hfill \\ ⟹\mathrm{\angle }DCF={38}^{\circ }+{22}^{\circ }={60}^{\circ },AF=CF\hfill \\ ⟹\mathrm{△}CDF\text{ is equilateral triangle}\hfill \\ ⟹\mathrm{\angle }CFD={60}^{\circ },CF=DF=AF\hfill \\ ⟹F\text{ is circumcenter of }\mathrm{△}ACD\hfill \\ ⟹\mathrm{\angle }CAD={\displaystyle \frac{\mathrm{\angle }CFD}{2}}={\displaystyle \frac{{60}^{\circ }}{2}}={30}^{\circ }\hfill \end{array}$

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2022/06/25

In quadrilateral $ABCD$, $\mathrm{\angle }CAD={30}^{\circ },\mathrm{\angle }CAB={20}^{\circ },\mathrm{\angle }ABD={30}^{\circ },\mathrm{\angle }CBD={10}^{\circ }$, find $\mathrm{\angle }BDC$.

Solve:

$\begin{array}{c}\text{Make }F\text{ on }AB\text{ so that }BC=CF,\text{Make }{D}^{\prime }\text{ on }AD\text{ so that }AF=F{D}^{\prime },\text{Connect }C{D}^{\prime },B{D}^{\prime }\hfill \\ ⟹\mathrm{\angle }CFB=\mathrm{\angle }CBF={30}^{\circ }+{10}^{\circ }={40}^{\circ },\mathrm{\angle }A{D}^{\prime }F=\mathrm{\angle }{D}^{\prime }AF={30}^{\circ }+{20}^{\circ }={50}^{\circ }\hfill \\ ⟹\mathrm{\angle }ACF=\mathrm{\angle }CFB-\mathrm{\angle }CAF={40}^{\circ }-{20}^{\circ }={20}^{\circ }=\mathrm{\angle }CAF,\mathrm{\angle }AF{D}^{\prime }={80}^{\circ }\hfill \\ ⟹AF=F{D}^{\prime }=CF=BC,\mathrm{\angle }CF{D}^{\prime }={180}^{\circ }-\mathrm{\angle }AF{D}^{\prime }-\mathrm{\angle }CFB={180}^{\circ }-{80}^{\circ }-{40}^{\circ }={60}^{\circ }\hfill \\ ⟹\mathrm{△}CF{D}^{\prime }\text{ is equilateral triangle}\hfill \\ ⟹C{D}^{\prime }=CF=CB\hfill \\ ⟹C\text{is circumcenter of }\mathrm{△}BF{D}^{\prime }\hfill \\ ⟹\mathrm{\angle }FB{D}^{\prime }={\displaystyle \frac{\mathrm{\angle }FC{D}^{\prime }}{2}}={\displaystyle \frac{{60}^{\circ }}{2}}={30}^{\circ }=\mathrm{\angle }FBD\hfill \\ ⟹D=D^{\prime }\hfill \\ ⟹\mathrm{\angle }BDC=\mathrm{\angle }B{D}^{\prime }C=\mathrm{\angle }CB{D}^{\prime }=\mathrm{\angle }CBD={10}^{\circ }\hfill \end{array}$

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