Friday, June 7, 2013

Problem 884: The Pythagorean Curiosity, Fifteen Conclusions

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 884.

Online Geometry Problem 884: The Pythagorean Curiosity, Fifteen Conclusions

1 comment:

  1. http://img442.imageshack.us/img442/5306/problem8841.png
    Partial answer to the problem.
    Questions 2, 5, 6 .
    note that ∆RCQ and ∆MXQ are congruent ( case SAS)
    So MQ=QR and ∠ (CQR)= ∠ (MQX)=y
    In isosceles triangle MQH we have ∠MQH)+2x=180
    At point Q we have ∠ (MQH)+2y+180=360 => ∠ (MQH)+2y=180
    So x=y and MH//XQ//AC
    Similarly DE//PY//BA
    Let Q’ is the projection of Q over QC and P’ is the projection of P over NB
    We have QQ’=CQ.sin(C )=b.c/a and PP’=PB.sin(B)=c.b/a
    So QQ’=PP’
    Note that QQ’=GG’ and PP’=FF’ ( congruent triangles)
    So FG//NR//BC
    Questions 9, 12, 13
    Note that right triangles XAY, MX’X and YY’D are congruent ( case HA)
    So MX’=AX= b => MH=4b
    So A’X’AY’ is a square and AA’ is a diagonal so AA’ is a bisector of angle A and A’ .
    AA’ ⊥ PQ

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