söndag, maj 22, 2011
This is my solution in response to "World's Second-Hardest Easy Geometry Problem".
Henceforth I use symbol ≅ to indicate triangle congruence.
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°
∠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.
Upplagd av Tsienni kl. 23:12