)nB  GSP!KP C`B +0 BCB4+0 A\B4+  AE)XKv(T| Problem: Given segment AB and point C, construct, using only a Euclidean (collapsing) compass, the image, C', of C under the translation that takes A to B. Thus one needs the point C' such that AB and CC' are parallel and equal in length. On an exam, one geometry student did this: Construct the equilateral triangle ACX (with X toward AB) and construct the equilateral triangle ABY (with Y toward C). Using X and Y construct the equilateral triangle XYH. Comparing the slope and length of AB with that of CH leads to the conjecture that H = C', the desired point. Draging C and B seems to confirm the conjecture. How can this be proven? Mel Thornton, University of Nebraska-LincolnlnlnW W0c7c0>LOt<! 4CB4C9?5%F?5%Ft! 3\B4C9?5%F?5%Fbl 2\B4BP ?5%F?5%F'9; 1`BBP ?5%F?5%F*P  n\B4`B?*0x  j\B4CB4?y $)nge GBm$) YBCM6_dR ExB X0A bX@  5 @88Length(Segment j) = p@HH! Z p@92t{.V 2.569 inches*A@@ CW  6 Slope(Segment j) = p@HH! Z p@92t{.V-0.000inches*A@@ *#*#  vCB4BCM6? *)"  u\B4BCM6? P&  t0A`B?0  s\B40A?`@ 50AC9?5%F?5%F lXl  6BCM6C9?5%F?5%F )X|  w0ABCM6? KP HC*BrwF J IC)J#y de  xC*BBCM6? P  y0AC*B?JPX|  z`BC*B?۬cW  7 @8qLength(Segment z) = p@HH! Z p@92t{.V 2.569 inches*A@@D 8 f#Slope(Segment z) = p@HH! Z p@92t{.V-0.000inches*A@@