söndag, maj 22, 2011

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,

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.

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