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Given a rectangle WXYZ, let R be a point on its circumscribed circle. Show that, out of the orthogonal projections of R onto WX, XY, YZ, and ZW; one out of these 4 points is the orthocenter of the triangle created by the other three.

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Have you tried saying $W,X,Y,Z$ are the points $(\pm a,\pm b)$ and $R=(r\cos\theta,r\sin\theta)$ for $r=\sqrt{a^2+b^2}$? Then the projections are $R_{1,2}=(\pm a,r\sin\theta)$ and $R_{3,4}=(r\cos\theta,\pm b)$ and it's fairly straightforward to check that each of the three pairs of lines $\{R_aR_b,R_cR_d\}$ are perpenticular. Thus each of the $R_i$ lies on the altitudes of the triangle formed by the other three points, proving the result. Remember that perpendicularity means the slopes are negative reciprocals, or that their products are $-1$. –  bgins Mar 19 '12 at 10:55

3 Answers 3

If you like complex numbers then the solution may be made very simple. WLOG we may assume that points $W, X, Y, Z$ all lie on the unit circle of the complex plane, and R is another point of the unit circle. We know that $WXYZ$ is a rectangle, so if, say, $W = a + i b$, then the remaining points are expressed as $X = \overline{W} = a - i b$, $Y = -W = - a - i b$, and $Z = - \overline{W} = -a + i b$.

Now let $R = p + i q$, and let us denote by $R_{A B}$ the projection of point $R$ onto line $AB$. Then by pure thought one can find that $R_{X Y} = p - i b$, $R_{Y Z} = - a + i q$, $R_{Z W} = p + i b$, and $R_{W X} = a + i q$.

We can now compute $R_{X Y} - R_{Y Z} = \overline{R} + \overline{W}$ and $R_{Z W} - R_{W X} = \overline{R} - \overline{W}$.

We need to observe that if points $A$ and $B$ lie on the unit circle, their sum $A + B$ and difference $A - B$ are orthogonal. Thus, line $(R_{X Y}, R_{Y Z})$ is perpendicular to line $(R_{Z W}, R_{W X})$

Finally, we find that line $(R_{X Y}, R_{Z W})$ is perpendicular to line $(R_{YZ}, R_{W X})$ by construction, so indeed point $R_{Z W}$ in the orthocenter of triangle $R_{XY}R_{YZ}R_{WX}$.

Similarly you may argue for the other points.

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Continuing with the notation and analytic-geometric approach from my comment, $$ \eqalign{ R_1R_2 \perp R_3R_4 \iff & \{y=r\sin\theta\} \perp \{x=r\cos\theta\} \\\\&\text{True (horizontal/vertical lines)} \\\\ R_1R_3 \perp R_2R_4 \iff & \overline{(a,r\sin\theta)(r\cos\theta,b)} \perp \overline{(-a,r\sin\theta)(r\cos\theta,-b)} \\\iff &-1= \frac{b-r\sin\theta}{r\cos\theta-a}\cdot\frac{-b-r\sin\theta}{r\cos\theta+a} =\frac{r^2\sin^2\theta-b^2}{r^2\cos^2\theta-a^2} \\&\qquad=\frac{a^2\sin^2\theta-b^2\cos^2\theta}{b^2\cos^2\theta-a^2\sin^2\theta} \\\\&\text{True} \\\\ R_1R_4 \perp R_2R_3 \iff & \overline{(a,r\sin\theta)(r\cos\theta,-b)} \perp \overline{(-a,r\sin\theta)(r\cos\theta,b)} \\\iff&-1= \frac{-b-r\sin\theta}{r\cos\theta-a}\cdot\frac{b-r\sin\theta}{r\cos\theta+a} \\\\&\text{...True?} } $$ Is this approach accesible for you? Can you try the last one? The facts we need are: $$ \matrix{ \cos^2\theta+\sin^2\theta=1\qquad&\text{true for all }\theta \\\\ a^2+b^2=r^2\qquad &\matrix{\text{do you see why I set}\\\text{it up this way above?}} \\\\ m_{PQ}=\frac{y_P-y_Q}{x_P-x_Q} &\text{slope formula} \\\\ -(s-t)=t-s &\text{basic algebra} \\\\ (s+t)(s-t)=s^2-t^2 &\text{basic algebra} \\\\ x=r\cos\theta &\text{is a circle with radius }r \\ y=r\sin\theta &\text{centered at the origin} } $$

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Synthetic solution:

Let the projections of $R$ onto $WX, ZY, WZ$, and $XY$ be $A,B,C,D$, respectively. (In my diagram $R$ lies on the arc $XY$, for what it's worth). Let $E$ be the intersection of line $AD$ with line $CB$.

Since trivially $CD \perp AB$ (intersecting at $R$), it's good enough to show that $ADE \perp CB$.

Claim 1: $RE \perp WY$. Proof: Consider the Simson line of $XWY$ with respect to the point $R$. The projections onto $XW$ and $AY$ are $A$ and $D$, so the projection onto $CB$ must be $E$.

Claim 2: $WARE$ is cyclic. Proof: $WA \perp RA$ and $RE \perp WE$.

Claim 3: $WARC$ is cyclic. Proof: $WA \perp AR$ and $WC \perp RC$.

Claim 4: $WAEC$ is cyclic. Proof: $WARE$ and $WARC$ are cyclic.

Claim 5: $AE \perp CE$. Proof: $WACE$ is cyclic and $WA \perp CW$.

Claim 6: $C,E,$ and $B$ are collinear. Proof: Simson line of $WZY$.

So $AE \perp CEB$, and $D$ is the orthocenter of $ABC$.

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