June 2022 Jun 07, 2022 2022/06/07 a,b,c>0,a32+b32+c32=1, prove: (1) abc≤19 (2) ab+c+bc+a+ca+b≤12abc prove: (1)G3=x1x2x33,A3=x1+x2+x33,Gn≤Ana32+b32+c32=1⟹a32+b32+c323=13⟹a32b32c323=(abc)12≤13⟹abc≤19◼(2)ab+c+bc+a+ca+b≤a2bc+b2ca+c2ab⟹ab+c+bc+a+ca+b≤a32+b32+c322abc=12abc◼ 2022/06/21 In △ABC, D is a point on BC so that ∠DAC=60∘,∠ACB=40∘,AC=BD, find ∠ABC. Solve: Easy to see ∠ADC=180∘−60∘−40∘=80∘Extend AD to E so that △ACE is equilateral.⟹AE=AC=CE,∠DCE=20∘Make F on BC so that CE=EF⟹CE=CF=AC=BD,∠CEF=∠CFE=80∘=∠ADC=∠FDE⟹BF+DF=CD+DF,∠FED=20∘,EF=DE,∠BFE=∠CDE⟹BF=CD⟹△BEF≅△CED⟹BE=CE=AE⟹E is the circumcenter of △ABC⟹∠ABC=∠AEC2=30∘ 2022/06/24 In quadrileteral ABCD, BC=CD,∠ABD=30∘,∠ACD=38∘,∠BAC=22∘, find ∠CAD. Solve: ∠CDB=∠BAC+∠ABD−∠ACD=14∘=∠CBD⟹∠BCD=180∘−2∗14∘−38∘=114∘Make F on AB so that CF=BC=CD⟹∠CFB=∠CBF=30∘+14∘=44∘⟹∠FCB=180∘−2∗44∘=92∘⟹∠ACF=114∘−92∘=22∘=∠BAC⟹∠DCF=38∘+22∘=60∘,AF=CF⟹△CDF is equilateral triangle⟹∠CFD=60∘,CF=DF=AF⟹F is circumcenter of △ACD⟹∠CAD=∠CFD2=60∘2=30∘ 2022/06/25 In quadrilateral ABCD, ∠CAD=30∘,∠CAB=20∘,∠ABD=30∘,∠CBD=10∘, find ∠BDC. Solve: Make F on AB so that BC=CF,Make D′ on AD so that AF=FD′,Connect CD′,BD′⟹∠CFB=∠CBF=30∘+10∘=40∘,∠AD′F=∠D′AF=30∘+20∘=50∘⟹∠ACF=∠CFB−∠CAF=40∘−20∘=20∘=∠CAF,∠AFD′=80∘⟹AF=FD′=CF=BC,∠CFD′=180∘−∠AFD′−∠CFB=180∘−80∘−40∘=60∘⟹△CFD′ is equilateral triangle⟹CD′=CF=CB⟹Cis circumcenter of △BFD′⟹∠FBD′=∠FCD′2=60∘2=30∘=∠FBD⟹D=D′⟹∠BDC=∠BD′C=∠CBD′=∠CBD=10∘ PREVIOUSMay 2022NEXTJuly 2022