Thursday, August 26, 2010

Problem 512: Triangle, Three Squares, Concurrency

Geometry Problem
Click the figure below to see the complete problem 512 about Triangle, Three Squares, Concurrency.

Problem 512: Triangle, Three Squares, Concurrency


See also:
Complete Problem 512

Level: High School, SAT Prep, College geometry

3 comments:

  1. Let P is the point of intersection of MB and CN. We will show that 3 points K,A and P are collinear.
    Tri ABC similar to tri. KMN with ratio BC/MN=AC/KN=PC/PN
    Note that angle (PCA)=angle (PNK) (corresponding angle)
    So tri. PCA similar to tri. PNK (case SAS)
    And angle(PAC)=angle(PKN) and 3 points P,A and K are collinear

    Peter Tran

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  2. To Peter

    It is not enough, two tr to be similar just by congruence of an angle
    Can you be more clearly, please

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  3. To c.t.e.o

    Consider 2 triangles PAC and PKN
    Angle PCA= Angle PNK ( corresponding angles)
    and PC/PN=BC/MN ( BC//MN and tri PBC similar to tri PMN))
    but BC/MN= AC/KN ( Tri. ABC similar to KMN)
    so PC/PN=AC/KN
    so tri. PAC similar to tri. PKN per case SAS
    Peter Tran

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