söndag, maj 22, 2011

World's Second-Hardest Easy Geometry Problem


This is my solution in response to "World's Second-Hardest Easy Geometry Problem".

Henceforth I use symbol ≅ to indicate triangle congruence.

Lemma:

In graph 1, suppose that triangle ΔIEH is equilateral, i.e. IE=EH=IH, and if ∠DIH=∠DHI=40°, then we conclude line DE bisects the angle ∠IEH, i.e., ∠DEI=∠DEH=(1/2)*∠IEH=(1/2)*60°=30°.

Proof: From point D, draw an altitude of the triangle ΔDIH which intersects line IH at point O. Since ∠DIH=∠DHI, and ∠DOI=∠DOH=90°, DO=DO, therefore triangle ΔDOI ≅ triangle ΔDOH, so IO=OH. Connects points O and E with a line OE, because IO=OH, OE=OE, IE=EH, therefore triangle ΔIOE ≅ triangle ΔHOE, so ∠IOE=∠HOE. Since ∠IOE+∠HOE=180°, it follows that ∠IOE=∠HOE=90°. So ∠DOH+∠HOE=180°, therefore points D, O, E lie on a straight line. Since line OE bisects the angle ∠IEH, we conclude that the same line DF bisects the angle ∠IEH, too.

Now back to the original problem. Draw a line HB counter-clockwise 20 degrees from line CB and line HB intersects line AE at point I, ∠IBA=∠IBD+∠DBA=∠CBD-∠CBI+∠DBA=30°-20°+50°=60°. Since∠IAB=60°, so triangle ΔIAB is equilateral, AB=AI=BI. Since ∠CAB=∠CBA=80°, triangle ΔCAB is isosceles, so AC=BC, ∠CAE=∠HBC=20°, ∠C=∠C, therefore triangle ΔCAE ≅ triangle ΔCBH, so CH=CE, triangle ΔCHE is also isosceles, line HE is parallel to line AB, ∠EHI=∠HEI=60°, hence triangle ΔHIE is also equilateral, therefore ∠HIE=60°.

Since ∠DAB=80°, ∠DBA=50°, ∠BDA=180°-80°-50°=50°, so triangle ΔADB is isosceles, AD=AB, but AB=AI, so AD=AI, therefore triangle ΔADI is isosceles, too. ∠DIA=(1/2)*(180°-∠DAI)=(1/2)*(180°-20°)=80°

Therefore ∠HID=180°-∠HIE-∠DIA=180°-60°-80°=40°

∠AHB=180°-∠HAB-∠HBA=180°-80°-60°=40°, i.e., ∠DHI=40°=∠HID, and we know triangle ΔHIE is equilateral. From the lemma we proved in the beginning, we conclude that line DE bisects the angle ∠AEH, i.e., ∠DEA=30°


Thus angle x is equal to 30 degrees.

World's Hardest Easy Geometry Problem


This is my solution in response to "World's Hardest Easy Geometry Problem".

Henceforth I use symbol ~ to denote being "similar" between two triangles; symbol ≅ to indicate triangle congruence.

Draw a line that bisects the angle ∠CBD - let it intersect line AE at point I and line AC at point P, so ∠CBP=∠PBD=(1/2)*20°=10°. Draw another line that is 10 degrees down clockwise from line AE and intersects line BD at point H. Because ∠IAB=70°, ∠IAH=10°, so ∠HAB=70°-10°=60°, and because ∠HBA=60°, so ∠BHA=180°-60°-60°=60°. Therefore triangle ΔABH is equilateral, meaning that AH=HB=AB.

Draw a line between point D and I. Because ∠DAI=∠IBD(=10°), it follows that the four points D, I, B, A lie on a circle, therefore the inscribed angle ∠IDB=∠IAB=70°, and the inscribed angle ∠DIA=∠DBA=60°. Line PDA and Line PIB are intersecting circle secants, therefore triangle ΔPDI ~ triangle ΔPBA, ∠PID=∠PAB=80°; ∠PDI=∠PBA=70°

and PI/DI=AP/AB (1)

Because ∠CAB=∠CBA=80° triangle ΔCAB is isosceles, CA=CB, likewise because ∠IAB=∠IBA=70° triangle ΔIAB is isosceles, IA=IB. Draw a line that connects points I and H, because AI=IB, and AH=HB, so triangle ΔAIH ≅ triangle ΔBIH, so ∠AIH=∠BIH=(1/2)*∠AIB=(1/2)*(180°-70°-70°)=20°. Draw a line that connects points C and I, because CA=CB, IA=IB, CI=CI, so triangle ΔCAI ≅ triangle ΔCBI, so ∠ACI=∠BCI=(1/2)*∠ACB=(1/2)*(180°-80°-80°)=10°. By symmetry both points I and H lie on the same line that bisects the angle ∠ACB. Also because ∠CAE=∠CBP=10°, ∠ACB=∠BCA, CA=CB, therefore triangle ΔCAE ≅ triangle ΔCPB, so PB=AE, since IA=IB, so PB-IB=AE=AI, ie, PI=EI. Then ∠PCI=∠ECI, PC=CE, CI=CI, so triangle ΔPCI ≅ triangle ΔECI, ∠CIP=∠CIE=∠APB-∠ACI=∠ACB+∠CBP-∠ACI=20°+10°-10°=20°.

Since ∠PDI=70°=∠IDH. Because ∠PID=80° and ∠HID=∠AIH+∠DIA=20°+60°=80°, so ∠PID=∠HID, DI=ID, so triangle ΔPDI ≅ triangle ΔHDI, therefore IH=IP=IE. Further, since ∠PCI=∠IBH=10°, ∠CIP=∠BIH=20°, and IP=IH, therefore triangle ΔCPI ≅ triangle ΔBHI so CE=CP=HB=HA=AB

Since triangle ΔPDI ~ triangle ΔPBA, and triangle ΔPDI ≅ triangle ΔHDI, so triangle ΔHDI and triangle ΔPBA would be similar as well,

so HI/DH=AP/PB
in other words, AP*DH=PB*HI (2)

Because ∠PCB=∠HIB=20°, and ∠CBP=∠IBH=10°, so triangle ΔPCB ~ triangle ΔHIB,
It follows that PB/PC=HB/HI, therefore PB*HI=CP*HB=HA*AB (3)

Compare (2) and (3) we have AP*DH=HA*AB, so HA/DH=AP/AB, but according to earlier conclusion (1), we know that AP/AB=PI/DI=EI/DI,

Therefore HA/DH=EI/DI
Since ∠DHI=180°-60°=120°, ∠DIE=∠DIP+∠CIP+∠CIE=80°+20°+20°=120°, so ∠DHI=∠DIE

Therefore triangle ΔDAH ~ triangle ΔDEI

Therefore ∠DEI=∠DAH=∠PAI+∠IAH=10°+10°=20°


Thus, angle x is equal to 20 degrees.

tisdag, maj 03, 2011

Horisont

Det fanns en tid som var varken vinter eller vår. Tiden, som bara hade strömmade ur de gamla tankarna, stod still denna isländska natt i början av maj månad. Under en mycket klar himmel troddes temperaturen utomhus ligga nära fryspunkten. När man kastade en blick åt blånande berg vid öppna havet längs horisonten, plötsligt kändes ens egen existens futtig och osäker. Sen började man erinra sig en världsbild vars innehåll berättade att allting kunde ju inte bli färdigt hädanefter. Den tiden försvann då, men jag fortfarande har dessa tankar i minnet.

Nu saknar jag Island ändå.