Question

Consider ADEF in the figure below.

The perpendicular bisectors of its sides are XW, YW, and ZW. They meet at a single point W.
(In other words, W is the circumcenter of ADEF.)
Suppose YW=32, DE=104, and FW=68.
Find EY, DW, and ZE.
Note that the figure is not drawn to scale.

Answer 1

Explanation:

Let's call the lengths of DW and EY as x and y, respectively.

Since W is the circumcenter of ADEF, we know that WY is the perpendicular bisector of DE, so WY = x/2.

Similarly, WZ is the perpendicular bisector of DF, so WZ = (104 - x) / 2.

Using the Pythagorean theorem, we can find the length of YE:

YE^2 = YW^2 + WZ^2

YE^2 = 32^2 + (104 - x)^2 / 4

YE^2 = 1024 + (104 - x)^2 / 4

Also, using the Pythagorean theorem, we can find the length of ZE:

ZE^2 = ZW^2 + WY^2

ZE^2 = 68^2 + x^2 / 4

ZE^2 = 4624 + x^2 / 4

Now, we have two equations with two unknowns, x and y.

Solving for y, we get:

y^2 = 1024 + (104 - x)^2 / 4

y^2 = 1024 + (104^2 - 208x + x^2) / 4

y^2 = 1024 + (104^2 - 208x + x^2) / 4

y^2 = 1024 + 26112 - 5216x + x^2 / 4

4y^2 = 26112 + x^2 - 208x + 4096

4y^2 = 26112 + x^2 - 208x + 4096

and

ZE^2 = 4624 + x^2 / 4

4ZE^2 = 18496 + x^2

Equating the two equations, we get:

4y^2 = 26112 + x^2 - 208x + 4096

= 18496 + x^2

7456 = 7616 - 208x

x = 104.

Finally,

DW = x = 104

EY = y = sqrt(1024 + (104^2 - 208x + x^2) / 4) = sqrt(1024 + (104^2 - 208 * 104 + 104^2) / 4) = sqrt(26112) = 162

ZE = ZW = 68.