Question

consider △def in the figure below. the perpendicular bisectors of its sides are \\(\overline{xw}\\), \\(\overline{yw}\\), and \\(\overline{zw}\\). they meet at a single point w. (in other words, w is the circumcenter of △def.) suppose \\(yw = 32\\), \\(dz = 64\\), and \\(fw = 68\\). find \\(ey\\), \\(dw\\), and \\(de\\). note that the figure is not drawn to scale.

Explanation:

Step1: Find EY

Since \( YW \) is the perpendicular bisector of \( EF \), \( EY = YF \). Also, in right triangle \( EYW \) or \( FYW \), but we can use the property of circumcenter: the perpendicular bisector of a side of a triangle passes through the circumcenter, so \( EY \) can be found using Pythagoras? Wait, no, wait. Wait, \( ZW \) is the perpendicular bisector of \( DE \), so \( DZ = ZE = 64 \), and \( XW \) is perpendicular bisector of \( DF \)? Wait, no, the problem says \( W \) is the circumcenter, so \( W \) is equidistant from \( D \), \( E \), \( F \), so \( DW = EW = FW \). Wait, \( FW = 68 \), so \( DW = 68 \)? Wait, no, let's re-examine.

Wait, \( YW \) is the perpendicular bisector of \( EF \), so \( EY = YF \), and \( \angle EYW = 90^\circ \). Also, \( ZW \) is the perpendicular bisector of \( DE \), so \( DZ = ZE = 64 \), so \( DE = DZ + ZE = 128 \)? Wait, no, \( Z \) is the midpoint of \( DE \), so \( DZ = ZE = 64 \), so \( DE = 2 \times 64 = 128 \).

Now, \( YW = 32 \), \( FW = 68 \). In right triangle \( FYW \), \( FY = \sqrt{FW^2 - YW^2} = \sqrt{68^2 - 32^2} \). Let's calculate that: \( 68^2 = 4624 \), \( 32^2 = 1024 \), so \( 4624 - 1024 = 3600 \), so \( \sqrt{3600} = 60 \). So \( FY = 60 \), and since \( YW \) is the perpendicular bisector, \( EY = FY = 60 \).

Step2: Find DW

Since \( W \) is the circumcenter, \( DW = FW \) (because circumcenter is equidistant from all vertices). Given \( FW = 68 \), so \( DW = 68 \).

Step3: Find DE

Since \( ZW \) is the perpendicular bisector of \( DE \), \( DZ = ZE = 64 \), so \( DE = DZ + ZE = 64 + 64 = 128 \).

Answer:

\( EY = 60 \), \( DW = 68 \), \( DE = 128 \)