USA TST 2014 #5
by Wolstenholme, Oct 15, 2014, 9:01 PM
Let 
be a cyclic quadrilateral, and let 
, 
, 
, and 
be the midpoints of 
, 
, 
, and 
respectively. Let 
, 
, 
and 
be the orthocenters of triangles 
, 
, 
and 
, respectively. Prove that the quadrilaterals and 
have the same area.
Proof:
We proceed with complex numbers. Let
have complex coordinates 
respectively and assume WLOG that the circumcircle of quadrilateral 
is the unit circle. Clearly 
and 
so the circumcenter of 
has coordinate 
. Therefore 
. Similarly 
and 
and 
.
Note![$ [ABCD] = [ABC] + [ACD] = \frac{i}{4}\left\lvert\begin{array}{ccc}a &\overline a & 1\\ b &\overline b & 1\\ c &\overline c & 1\\ \end{array}\right\rvert + \frac{i}{4}\left\lvert\begin{array}{ccc}d &\overline d & 1\\ a &\overline a & 1\\ c &\overline c & 1\\ \end{array}\right\rvert $](//latex.artofproblemsolving.com/0/5/2/052c658891b84b73683370943caf85695484c86a.png)
.
Similarly,![$ [WXYZ] = [WXY] + [WZY] = \frac{i}{4}\left\lvert\begin{array}{ccc} \frac{2a + b + d}{2} & \frac{2bd + ad + ab}{2abd} & 1\\ \frac{2b + a + c}{2} & \frac{2ac + ba + bc}{2abc} & 1\\ \frac{2c + b + d}{2} & \frac{2bd + cd + cb}{2cbd} & 1\\ \end{array}\right\rvert - $](//latex.artofproblemsolving.com/3/0/1/3018a89a138221e954df5e6c79414e9c0f4692b3.png)
Which after some (surprisingly easy) computation comes out to
as desired.
be a cyclic quadrilateral, and let 
, 
, 
, and 
be the midpoints of 
, 
, 
, and 
respectively. Let 
, 
, 
and 
be the orthocenters of triangles 
, 
, 
and 
, respectively. Prove that the quadrilaterals and 
have the same area.Proof:
We proceed with complex numbers. Let
have complex coordinates 
respectively and assume WLOG that the circumcircle of quadrilateral 
is the unit circle. Clearly 
and 
so the circumcenter of 
has coordinate 
. Therefore 
. Similarly 
and 
and 
.Note
![$ [ABCD] = [ABC] + [ACD] = \frac{i}{4}\left\lvert\begin{array}{ccc}a &\overline a & 1\\ b &\overline b & 1\\ c &\overline c & 1\\ \end{array}\right\rvert + \frac{i}{4}\left\lvert\begin{array}{ccc}d &\overline d & 1\\ a &\overline a & 1\\ c &\overline c & 1\\ \end{array}\right\rvert $](http://latex.artofproblemsolving.com/0/5/2/052c658891b84b73683370943caf85695484c86a.png)
.Similarly,
![$ [WXYZ] = [WXY] + [WZY] = \frac{i}{4}\left\lvert\begin{array}{ccc} \frac{2a + b + d}{2} & \frac{2bd + ad + ab}{2abd} & 1\\ \frac{2b + a + c}{2} & \frac{2ac + ba + bc}{2abc} & 1\\ \frac{2c + b + d}{2} & \frac{2bd + cd + cb}{2cbd} & 1\\ \end{array}\right\rvert - $](http://latex.artofproblemsolving.com/3/0/1/3018a89a138221e954df5e6c79414e9c0f4692b3.png)
Which after some (surprisingly easy) computation comes out to
as desired.
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