August 20th, 2017, 03:10 PM  #1 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,593 Thanks: 937 Math Focus: Elementary mathematics and beyond  Geometry Challenge Prove that the three altitudes of a triangle intersect at a single point. Last edited by greg1313; August 25th, 2017 at 12:19 AM. 
September 2nd, 2017, 04:48 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,729 Thanks: 705 
We can always set up a coordinate system such that the vertices of the triangle are at (0, 0), (a, 0), and (b, c). One side is the horizontal line y= 0 and the altitude on that side is x= b. Another side is y= cx/b. Any line perpendicular to that has slope b/c. The altitude on that side must pass through (a, 0) so has equation y= b(x a)/c. The third side has equation y= c(x a)/(b a) so has slope c/(b a). Any line perpendicular to that has slope (a b)/c. The altitude on that side must go through (0, 0) so has equation y= (a b)x/c. Show that there exist a unique (x, y) satisfying x= b. y= b(x a)/c, and y= (a b)x/c. 
September 6th, 2017, 03:46 PM  #3 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 561 Thanks: 79 
When x = b, the last two equation are the same. In the second equation, change x to b to make y = b(b  a)/c. In the third equation, change x to b and put the b in front to make y =b(a  b)/c. Multiplying the parenthesis and changing the order of the terms makes both equation become y = (ab  b^2)/c.


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