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


  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

  2. To Peter

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

  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