very easy problem! We get that $AW=CY$, $AW\parallel CY$ $ \Rightarrow $ $AWYC$-parallelogram. Similarly, $BXZD$ is parallelogram. Then $WY=AC$, $BD=XZ$, $WY\parallel AC$, $BD\parallel XZ$ $ \Rightarrow $ $ [ABCD]=[WXYZ] $.

Yes, apply rotational geometry to see that $AW = CY$ and $BX = DZ$.

such an easy problem for USA team selection ? seems weird

We proceed with complex numbers. Let $ A, B, C, D, E, F, G, H, W, X, Y, Z $ have complex coordinates $ a, b, c, d, e, f, g, h, w, x, y, z $ respectively and assume WLOG that the circumcircle of quadrilateral $ ABCD $ is the unit circle. Clearly $ h = \frac{a + d}{2} $ and $ e = \frac{a + b}{2} $ so the circumcenter of $ \triangle{AHE} $ has coordinate $ \frac{a}{2} $. Therefore $ w = a + h + e - 2\left(\frac{a}{2}\right) = a + \frac{b + d}{2} $. Similarly $ y = c + \frac{b + d}{2} $ and $ x = b + \frac{a + c}{2} $ and $ z = d + \frac{a + c}{2} $. 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 $ $ = \frac{i}{4}\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} - \frac{b}{a} - \frac{c}{b} - \frac{d}{c} - \frac{a}{d}\right) $. 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 - $ $ \frac{i}{4}\left\lvert\begin{array}{ccc} \frac{2a + b + d}{2} & \frac{2bd + ad + ab}{2abd} & 1\\ \frac{2d + a + c}{2} & \frac{2ac + da + dc}{2adc} & 1\\ \frac{2c + b + d}{2} & \frac{2bd + cd + cb}{2cbd} & 1\\ \end{array}\right\rvert $ Which after some (surprisingly easy) computation comes out to $ \frac{i}{4}\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} - \frac{b}{a} - \frac{c}{b} - \frac{d}{c} - \frac{a}{d}\right) $ as desired.

Let $H_a$ be the orthocenter of $\triangle BCD$ and define $H_b, H_c, H_d$ similarly. By considering the homothety $\mathcal{H}(A, 2)$, we deduce that $W$ is the midpoint of $\overline{AH_a}.$ Similar relations hold for the other vertices. Now, let $\mathcal{C}$ be the rectangular hyperbola passing through $A, B, C, D.$ It is well-known that $\mathcal{C}$ passes through $H_a$, and that the center $O$ of $\mathcal{C}$ is the midpoint of $\overline{AH_a}.$ Similar relations hold for the other vertices, so it follows from the homothety $\mathcal{H}(O, -1)$ that $ABCD \sim H_aH_bH_cH_d.$ Thus from the Mean Geometry Theorem, we have $WXYZ \sim ABCD.$ Meanwhile, since $ABH_aH_b$ is a parallelogram (because its diagonals bisect one another), we have $AB = H_aH_b = WX.$ Therefore, $WXYZ \cong ABCD$, and the desired result follows. $\square$

We use complex numbers

As a side note, how much work does one need to show in a complex bash during the olympiad? Do you need to write out all your expansions, etc.?

Let the vertices be $a$, $b$, $c$, and $d$. Through direct computation we have $$e=\frac{a+b}{2} \;\;\; f=\frac{b+c}{2} \;\;\; g=\frac{c+d}{2} \;\;\; h=\frac{a+d}{2}$$ By considering homotheties of scale factor 2 mapping triangles $\triangle{AHE}$, $\triangle{BEF}$, $\triangle{CFG}$ and $\triangle{DGH}$ to $\triangle{DAB}$, $\triangle{ABC}$, $\triangle{BCD}$, and $\triangle{CDA}$ respectively, we see that $$w=\frac{d+2a+b}{2} \;\;\; x=\frac{a+2b+c}{2} \;\;\; y=\frac{b+2c+d}{2} \;\;\; z=\frac{a+c+2d}{2}$$ We scale areas down by a factor of $\frac{i}{4}$, before we compute. $[WXYZ]=[WZY]+[ZYX]$, which is equal to $$\begin{vmatrix} z& \overline{z}& 1\\ w& \overline{w}& 1\\ y& \overline{y}& 1\\ \end{vmatrix}+\begin{vmatrix} x& \overline{x}& 1\\ y& \overline{y}& 1\\ w& \overline{w}& 1\\ \end{vmatrix}=\begin{vmatrix} z& \overline{z}& 1\\ w& \overline{w}& 1\\ y& \overline{y}& 1\\ \end{vmatrix}-\begin{vmatrix} x& \overline{x}& 1\\ w& \overline{w}& 1\\ y& \overline{y}& 1\\ \end{vmatrix}=\begin{vmatrix} z-x& \overline{z-x}& 0\\ w& \overline{w}& 1\\ y& \overline{y}& 1\\ \end{vmatrix}$$Plugging in $w$, $x$, $y$, and $z$, and simplifying, we have $$\begin{vmatrix} (d-b)& \overline{(d-b)}& 0\\ \frac{2a+b+d}{2} & \overline{\frac{2a+b+d}{2}}& 1\\ \frac{2c+b+d}{2} & \overline{\frac{2c+b+d}{2}}& 1\\ \end{vmatrix}=\frac{1}{4}\begin{vmatrix} 2(d-b)& \frac{-2(d-b)}{bd}& 0\\ 2a& \frac{2}{a}& 1\\ 2c& \frac{2}{c}& 1\\ \end{vmatrix}=\begin{vmatrix} (d-b)& -\frac{(d-b)}{bd}& 0\\ a& \frac{1}{a}& 1\\ c& \frac{1}{c}& 1\\ \end{vmatrix}$$$[ABCD]=[ABC]+[ACD]$, which is equal to $$\begin{vmatrix} a& \overline{a}& 1\\ b& \overline{b}& 1\\ c& \overline{c}& 1\\ \end{vmatrix}+\begin{vmatrix} a& \overline{a}& 1\\ c& \overline{c}& 1\\ d& \overline{d}& 1\\ \end{vmatrix}=\begin{vmatrix} d& \overline{d}& 1\\ a& \overline{a}& 1\\ c& \overline{c}& 1\\ \end{vmatrix}-\begin{vmatrix} b& \overline{b}& 1\\ a& \overline{a}& 1\\ c& \overline{c}& 1\\ \end{vmatrix}=\begin{vmatrix} (d-b)& -\frac{(d-b)}{bd}& 0\\ a& \frac{1}{a}& 1\\ c& \frac{1}{c}& 1\\ \end{vmatrix}$$. Hence $[WXYZ]=[ABCD]$ as desired. $\square$

Delray wrote: Plugging in $w$, $x$, $y$, and $z$, and simplifying, we have $$\begin{vmatrix} (d-b)& \overline{(d-b)}& 0\\ \frac{2a+b+d}{2} & \overline{\frac{2a+b+d}{2}}& 1\\ \frac{2c+b+d}{2} & \overline{\frac{2c+b+d}{2}}& 1\\ \end{vmatrix}=\frac{1}{4}\begin{vmatrix} 2(d-b)& \frac{-2(d-b)}{bd}& 0\\ 2a& \frac{2}{a}& 1\\ 2c& \frac{2}{c}& 1\\ \end{vmatrix}=\begin{vmatrix} (d-b)& -\frac{(d-b)}{bd}& 0\\ a& \frac{1}{a}& 1\\ c& \frac{1}{c}& 1\\ \end{vmatrix}$$ Can someone explain me how to get the middle part? I am new to complex bashing and detrminants :/

There is a diagram https://drive.google.com/file/d/1-rBBU2vvCoxIBmtN_5h6i48_AnlcSZAu/view?usp=drivesdk

I claim that $WAYC$ and $XBZD$ are parallelograms. Clearly, $WA\parallel YC$, since they are both perpendicular to $BD$. Supposing $\angle A$ is acute, we get that $WA=\frac{AH\cos A}{\sin\angle HEA}=R\cos A$. Likewise, $YC=-R\cos C$, so $WA=YC$, and $WAYC$ is a parallelogram as desired. Similarly, we get $XBZD$ is a parallelogram. Now, note that the area of a quadrilateral is $\frac{1}{2}ab\sin\phi$, where $a,b$ are the lengths of its diagonals, and $\phi$ is the angle between them. We have that $WY=AC$, $XZ=BD$, and the angle between $WY,XZ$ is the same as that between $AC,BD$ (parallel), so $[ABCD]=[WXYZ]$, as desired.

Observe that \[\overrightarrow{YW}=\frac{\overrightarrow{OA}+\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OD}\right)}{2}-\frac{\overrightarrow{OC}+\left(\overrightarrow{OC}+\overrightarrow{OD}+\overrightarrow{OB}\right)}{2}=\overrightarrow{OA}-\overrightarrow{OC}=\overrightarrow{CA}\]and similarly $\overrightarrow{ZX}=\overrightarrow{DB}$ so by cross product, $ABCD$ and $WXYZ$ have the same area.

Let $H_1$ and $H_2$ be the orthocenters of triangles $ABC$ and $BCD$. Then we see that $AW=CY$, also, as $AW$ is parallel to $CY$ it follows that $AWYC$ is a parallelogram and the result follows. It's also easy to bash with complex numbers.

Toss onto the complex plane. Note that a homothety at $A$ with scale factor 2 sends $AHE \to ADB$, so $w-a = \tfrac12(a+b+d-a) \implies w = a+\tfrac{b}{2} + \tfrac{d}{2}$. We want to show that the area of the quad formed by \[ w = a+\tfrac{b}{2}+\tfrac{d}{2}, x=b+\tfrac{c}{2} + \tfrac{a}{2}, y = c + \tfrac{d}{2} + \tfrac{b}{2}, z = d+\tfrac{a}{2} + \tfrac{c}{2} \]is the same as the area of $ABCD$. Translate $WXYZ$ by $-\tfrac{a+b+c+d}{2}$, so it becomes \[ w' = \tfrac{a-c}{2}, x' = \tfrac{b-d}{2}, y' = \tfrac{c-a}{2}, z' = \tfrac{d-b}{2}. \]Note that this does not change the area. This new quad is a parallelogram, since the midpoint of $WY$ and $XZ$ is $O$. Here, $W'Y'=|a-c|=AC$ and $X'Z'=BD$. So the area of $W'X'Y'Z'$ is $\tfrac12 AC\cdot BD \sin \theta$, where $\theta$ is the angle between $AC$ and $BD$ (using area of parallelogram rule). But this is just the area of $ABCD$ (use the $\tfrac12 ab\sin C$ formula 4 times for each of the 4 triangles formed by cutting diagonals), so we are done.

oops this is literally the same as the above solution but anyway posting for storage Set $(ABCD)$ as the unit circle. The homothety about $A$ with factor $2$ sends $W$ to the orthocenter of $\triangle{ABD}$, which is $a+b+d$; thus, $w=a+\frac{b+d}{2}$, and similarly we have $x=b+\frac{a+c}{2}, y=c+\frac{b+d}{2}$, and $z=d+\frac{a+c}{2}$. But note that $w-y=a-c$ and $x-z=b-d$, which finishes the problem due to the formula $[PQRS]=\frac{1}{2}(PR)(QS)\sin\angle{(PQ, RS)}$.

We dump it on the complex plane. Let $A,B,C,D,W,X,Y,Z$ have coordinates $a,b,c,d,w,x,y,z$ respectively and let the circumcenter of $ABCD$ be the origin. Note that $E,F,G,H$ have coordinates $\frac{a+b}{2},\frac{b+c}{2}, \frac{c+d}{2},\frac{a+d}{2}$. By homothety, the midpoint of the orthocenter of $ABD$ and $AEH$ is $a$. Similar arguments hold for the other triangles. It follows that $w,x,y,z$ are $\frac{2a+b+d}{2},\frac{2b+c+a}{2},\frac{2c+a+b}{2},\frac{2d+c+a}{2}$ respectively. At this point, calculating the area of $WXYZ$ seems daunting, but we can translate $WXYZ$ down by $a+b+c+d$(forming $W’X’Y’Z’$) to get: $w’ = \frac{a-c}{2},x’ = \frac{b-d}{2},y’=\frac{c-a}{2},z’ = \frac{d-b}{2}$. The area is equal to, noting $WXYZ$ is a parallelogram: \begin{align*} [WXYZ] &= [W’X’Y’Z’] \\ &= [W’X’Y’] + [W’Y’Z’] \\ &= 2 [W’X’Y’] \\ &= \frac{i}{4}\begin{vmatrix} \frac{a-c}{2}& \frac{\overline{a-c}}{2}& 1\\ \frac{b-d}{2}& \frac{\overline{b-d}}{2}& 1\\ \frac{c-a}{2}& \frac{\overline{c-a}}{2}& 1\\ \end{vmatrix} \\ &= \frac{i}{2} \left( \frac{(a-c)(\bar b - \bar d) - (\bar a - \bar c)(b-d)}{4} + \frac{(b-d)(\bar c - \bar a) - (\bar b - \bar d)(c-a)}{4} \right) \\ &= \frac{i}{2} \left(\frac{(a-c)(\bar b - \bar d) - (\bar a - \bar c)(b-d)}{2} \right)\end{align*}In addition, \begin{align*} [ABCD] &= [ABC] + [ACD] \\ &= \begin{vmatrix} a & \bar a & 1 \\ b & \bar b & 1 \\ c & \bar c & 1 \\ \end{vmatrix} + \begin{vmatrix} a & \bar a & 1 \\ c & \bar c & 1 \\ d & \bar d & 1 \\ \end{vmatrix} \\ &= \frac{i}{2} \left(\frac{a \bar b - b \bar a + b \bar c - c \bar b - a \bar c + c \bar a + a \bar c - \bar a c + c \bar d - \bar c d - a \bar d + \bar a d}{2} \right) \\ &= \frac{i}{2} \left(\frac{a \bar b - b \bar a + b \bar c - c \bar b + c \bar d - \bar c d - a \bar d + \bar a d}{2} \right) \end{align*} from which we verify that $[ABCD] = [WXYZ]$.

Free variables are $a,b,c,d$, all on the unit circle. Note that a homothety with factor 2 at A takes $W$ to the orthocenter of $ABD$. Thus, we have $w = \frac{(a)+(d+a+b)}{2}=\frac{d+2a+b}{2}$. By similar logic, $x=\frac{a+2b+c}{2},y=\frac{b+2c+d}{2}, z=\frac{c+2d+a}{2}$. We can shift it by $\frac{a+b+c+d}{2}$ to get $w'=\frac{a-c}{2},x'=\frac{b-d}{2},y'=\frac{c-a}{2},z'=\frac{d-b}{2}$. We now note that \[[WXYZ]=[W'X'Y'Z'] = 4\cdot [\triangle OW'X'] = 4\cdot (\frac{1}{4}\left|(\frac{a-c}{2})(\frac{\frac1b-\frac1d}{2})-(\frac{\frac1a-\frac1c}{2})(\frac{b-d}{2})\right|) =\]\[=\frac{1}{4}\left[(\frac{a}{b}-\frac{c}{b}-\frac{a}{d}+\frac{c}{d})-(\frac{b}{a}-\frac{b}{c}-\frac{d}{a}+\frac{d}{c}) \right]\] Compare that with \[ABCD = AOB+BOC+COD+DOA = \frac{1}{4}(a\cdot b'-\frac{1}{a}b)+\frac{1}{4}(b/c-c/b)+\frac{1}{4}(c/d-d/c)+\frac{1}{4}(d/a-a/d) \]This is clearly equal to $[WXYZ]$ so we are done $\blacksquare$.

Let $a,b,c,d$ be on the unit circle. We will compute $w,x,y,z$ first. Observe that a homothety of scale factor $2$ centered at $A$ sends $W$ to the orthocenter of $\triangle ADB$, which is just $a+b+d$ as $(ADB)$ is the unit circle. Hence we have $w=\tfrac{1}{w}(a+b+d-a)+a=a+\tfrac{b+d}{2}$. Similarly, we obtain $x=b+\tfrac{c+a}{2},y=c+\tfrac{d+b}{2},z=d+\tfrac{a+c}{2}$. Now we want to show that $[ABCD]=[WXYZ] \iff [ABC]+[CDA]=[WXY]+[YZW]$, which by complex shoelace occurs only if: $$\begin{vmatrix}a&\overline{a}&1\\ b&\overline{b}&1\\c&\overline{c}&1\end{vmatrix}+\begin{vmatrix}c&\overline{c}&1\\ d&\overline{d}&1\\a&\overline{a}&1\end{vmatrix}=\begin{vmatrix}w&\overline{w}&1\\ x&\overline{x}&1\\y&\overline{y}&1\end{vmatrix}+\begin{vmatrix}y&\overline{y}&1\\ z&\overline{z}&1\\w&\overline{w}&1\end{vmatrix}.$$It's not difficult to explicitly compute (without any "shortcuts"): $$\begin{vmatrix}a&\overline{a}&1\\ b&\overline{b}&1\\c&\overline{c}&1\end{vmatrix}+\begin{vmatrix}c&\overline{c}&1\\ d&\overline{d}&1\\a&\overline{a}&1\end{vmatrix}=\frac{a}{b}-\frac{b}{a}+\frac{b}{c}-\frac{c}{b}+\frac{c}{d}-\frac{d}{c}+\frac{d}{a}-\frac{a}{d},$$where we use the fact that $a=\tfrac{1}{a}$ as $a$ lies on the unit circle (and similar facts for $b,c,d$). Now, note that $\overline{w}=\tfrac{1}{a}+\tfrac{1}{2b}+\tfrac{1}{2d}$, and similar identities hold for the other three variables. Hence: \begin{align*} \begin{vmatrix}w&\overline{w}&1\\ x&\overline{x}&1\\y&\overline{y}&1\end{vmatrix}+\begin{vmatrix}y&\overline{y}&1\\ z&\overline{z}&1\\w&\overline{w}&1\end{vmatrix}&=\begin{vmatrix}w&\overline{w}&1\\ x&\overline{x}&1\\y&\overline{y}&1\end{vmatrix}-\begin{vmatrix}w&\overline{w}&1\\ z&\overline{z}&1\\y&\overline{y}&1\end{vmatrix}\\ &=\begin{vmatrix}w&\overline{w}&1\\ x-z&\overline{x-z}&0\\y&\overline{y}&1\end{vmatrix}\\ &=\begin{vmatrix}a+\frac{b}{2}+\frac{d}{2}&\frac{1}{a}+\frac{1}{2b}+\frac{1}{2d}&1\\ b-d&\frac{1}{b}-\frac{1}{d}&0\\c+\frac{d}{2}+\frac{b}{2}&\frac{1}{c}+\frac{1}{2b}+\frac{1}{2d}&1\end{vmatrix}\\ &=\begin{vmatrix}a+\frac{b}{2}+\frac{d}{2}&\frac{1}{a}+\frac{1}{2b}+\frac{1}{2d}&1\\ b-d&\frac{1}{b}-\frac{1}{d}&0\\c-a&\frac{1}{c}-\frac{1}{a}&0\end{vmatrix}\\ &=\begin{vmatrix}b-d&\frac{1}{b}-\frac{1}{d}\\ c-a&\frac{1}{c}-\frac{1}{a}\end{vmatrix}\\ &=\frac{a}{b}-\frac{b}{a}+\frac{b}{c}-\frac{c}{b}+\frac{c}{d}-\frac{d}{c}+\frac{d}{a}-\frac{a}{d}, \end{align*}so $[ABCD]=[WXYZ]$ as desired. $\blacksquare$

I wish I was alive in 2014 $W$ is the image of the orthocenter of $ADB$ under the homothety with scale factor $\frac{1}{2}$ centered at $A$. In complex numbers, it's hence equal to $\frac{2a+b+d}{2}$ and symmetrically for the other three orthocenters. Now, notice that $w-y=a-c$ and $x-z=b-d$, which immediately kills by applying absinc/2 four times.

mathleticguyyy wrote: I wish I was alive in 2014 You weren't??

WLOG, assume $\angle B \ge 90^{\circ}$, suppose $ABCD$ is convex, and let $H_a, H_b, H_c, H_d$ be the orthocenters of $BCD, CDA, DAB, ABC$ respectively. It's well-known that the midpoints of $AH_a, BH_b, CH_c, DH_d$ each coincide with the Euler-Poncelet Point of $ABCD$, which we denote by $P$. Now, it's easy to see $ACH_aH_c$ and $BDH_bH_d$ are parallelograms. In addition, homotheties imply $W, X, Y, Z$ are the midpoints of $AH_c, BH_d, CH_a, DH_b$ respectively. Thus, properties of parallelograms and midlines yield $$AC = WY = H_cH_a, \hspace{0.2mm} AC \parallel WY \parallel H_cH_a$$and $$BD = XY = H_dH_b, \hspace{0.2mm} BD \parallel XY \parallel H_dH_b.$$Now, it's clear $$\text{dist}(B, AC) + \text{dist}(D, AC)$$$$= \text{dist}(B, AC) + \left( DZ + \text{dist}(Z, AC) \right)$$$$= \text{dist}(B, AC) + BX + \text{dist}(Z, AC)$$$$= \text{dist}(X, AC) + \text{dist}(Z, AC) = \text{dist}(X, WY) + \text{dist}(Z, WY)$$where the last equality is true because $AC$ and $WY$ are parallel and neither of $X, Z$ are contained between lines $AC$ and $WY$. Thus, we know $$[ABCD] = \frac{\text{dist}(B, AC) + \text{dist}(D, AC)}{2} \cdot AC$$$$= \frac{\text{dist}(X, WY) + \text{dist}(Z, WY)}{2} \cdot WY = [WXYZ]$$as desired. $\blacksquare$ Remarks: If $\angle B > 90^{\circ}$, then $$\text{dist}(H_d, AC) > \text{dist}(B, AC).$$In other words, the purpose of our initial assumption is to eradicate configuration issues. If $\angle B = 90^{\circ}$, then $H_d$ coincides with $B$, which trivializes this question. Once we find the two triplets of pairwise congruent and parallel segments, vectors can be used to easily show $[ABCD] = [WXYZ].$ (One can also construct congruent triangles.) Unfortunately, I suck at using vectors formally and was forced to bash with altitudes. Also, $WXYZ$ is actually a parallelogram whose diagonals meet at $P$. (This result follows easily from homotheties.)

mathleticguyyy wrote: I wish I was alive in 2014 $W$ is the image of the orthocenter of $ADB$ under the homothety with scale factor $\frac{1}{2}$ centered at $A$. In complex numbers, it's hence equal to $\frac{2a+b+d}{2}$ and symmetrically for the other three orthocenters. Now, notice that $w-y=a-c$ and $x-z=b-d$, which immediately kills by applying absinc/2 four times. doing tst problems as a preschooler how so orz

Just show that $\triangle WHZ\cong \triangle XFY$ and $\triangle WXE\cong \triangle ZYG$ and finally that $AW=CY$ and $BX=DZ$ and done.

We will use complex numbers. Let $(ABCD)$ be the unit circle. Note that $e=\frac{a+b}{2}, f=\frac{b+c}{2}, g=\frac{c+d}{2}, h=\frac{a+d}{2}$. We have $\triangle{AHE}$ is a dilation of scale factor $\frac{1}{2}$ from $\triangle{ADB}$ centered at point $A$. Note that the orthocenter of $\triangle{ADB}$ is $a+b+d$. So we subtract by $a$, divide by $2$, and add $a$ again, to get the orthocenter of $AHE$. $w=a+\frac{b+d}{2}$. Similarly, $x=b+\frac{a+c}{2}, y=c+\frac{b+d}{2}, z=d+\frac{a+c}{2}$. Note that \[[ABCD]=[ABC]+[ACD]=\frac{i}{4}\left(\begin{vmatrix} a& \overline{a}& 1\\ b& \overline{b}& 1\\ c& \overline{c}&1\\ \end{vmatrix}+\begin{vmatrix} a& \overline{a}& 1\\ c& \overline{c}&1\\ d& \overline{d}&1\\ \end{vmatrix}\right)=\frac{i}{4}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}-\frac{b}{a}-\frac{c}{b}-\frac{d}{c}-\frac{a}{d}\right) \] Note that \[[WXYZ]=[WXY]+[WYZ]=\frac{i}{4}\left(\begin{vmatrix} w& \overline{w}& 1\\ x& \overline{x}& 1\\ y& \overline{y}&1\\ \end{vmatrix}+\begin{vmatrix} w& \overline{w}& 1\\ y& \overline{y}&1\\ z& \overline{z}&1\\ \end{vmatrix}\right)= \frac{i}{4}\left(\begin{vmatrix} \frac{2a+b+d}{2}& \frac{2bd+ab+ad}{2abd}& 1\\ \frac{2b+a+c}{2}& \frac{2ac+ab+bc}{2abc}& 1\\ \frac{2c+b+d}{2}& \frac{2bd+cb+cd}{2bcd}&1\\ \end{vmatrix}+\begin{vmatrix} \frac{2a+b+d}{2}& \frac{2bd+ab+ad}{2abd}& 1\\ \frac{2c+b+d}{2}& \frac{2bd+cb+cd}{2bcd}&1\\ \frac{2d+a+c}{2}& \frac{2ac+da+dc}{2acd}&1\\ \end{vmatrix}\right)\] A bunch of terms here cancel out. For simplicity, divide by $\frac{i}{4}$ because we can multiply by that at the end. We get \[\frac{2a+b+d}{2}\left(\frac{2ac+ab+bc}{2abc}-\frac{2ac+da+dc}{2acd}\right)+\frac{2bd+ab+ad}{2abd}\left(\frac{2d+a+c}{2}-\frac{2b+a+c}{2}\right)+\frac{2ac+ab+bc}{2abc}-\frac{2ac+da+dc}{2acd}.\] \[\begin{vmatrix} x_1& x_2& 1\\ x_3& x_4& 1\\ x_5& x_6&1\\ \end{vmatrix}+\begin{vmatrix} x_1& x_2& 1\\ x_5& x_6&1\\ x_7& x_8&1\\ \end{vmatrix}=x_1x_4-x_1x_6-x_2x_3+x_2x_5+x_3x_6-x_4x_5+x_1x_6-x_1x_8-x_2x_5+x_2x_7+x_5x_8-x_6x_7=x_1x_4-x_2x_3+x_3x_6-x_4x_5-x_1x_8+x_2x_7+x_5x_8-x_6x_7\] But we note $x_7=x_1+x_5-x_3$ and $x_8=x_2+x_6-x_4$. So the sum of determinants is \[x_1x_4-x_2x_3+x_3x_6-x_4x_5-x_1x_2-x_1x_6+x_1x_4+x_2x_1+x_2x_5-x_2x_3+x_5x_2+x_5x_6-x_5x_4-x_6x_1-x_6x_5+x_6x_3=2x_1x_4-2x_2x_3+2x_3x_6-2x_4x_5-2x_1x_6+2x_2x_5=x_4(2x_1-2x_5)+x_2(2x_5-2x_3)+x_6(2x_3-2x_1)=x_4(2a-2c)+x_2(c+d-b-a)+x_6(b+c-a-d).\] Note that \[x_4=\frac{2ac+ab+bc}{2abc}=\frac{1}{b}+\frac{1}{2a}+\frac{1}{2c}, x_2=\frac{1}{a}+\frac{1}{2b}+\frac{1}{2d}, x_6=\frac{1}{c}+\frac{1}{2b}+\frac{1}{2d}.\] So sum is equal to \[\frac{2a}{b}-\frac{2c}{b}+1-\frac{2c}{2a}+\frac{2a}{2c}-1+\frac{c}{a}+\frac{d}{a}-\frac{b}{a}-1+\frac{b}{c}+1-\frac{a}{c}-\frac{d}{c}+\left(\frac{1}{2b}+\frac{1}{2d}\right)\left(2c-2a\right)=\frac{2a}{b}-\frac{2c}{b}-\frac{c}{a}+\frac{a}{c}+\frac{c}{a}+\frac{d}{a}-\frac{b}{a}+\frac{b}{c}-\frac{a}{c}-\frac{d}{c}+\frac{c}{b}-\frac{a}{b}+\frac{c}{d}-\frac{a}{d}=\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}-\frac{b}{a}-\frac{c}{b}-\frac{d}{c}-\frac{a}{d},\]which proves the areas are equal.

POV: you started to complex but found a way out of it because you aren't a true bash main Let $M$ be the midpoint of $BD$. Claim: $WA=OM$ and $WA\parallel OM$. We will use complex bash. Let $(ABCD)$ be the unit circle. Note that the orthocenter of $ABD$ is $a+b+d$, so the orthocenter of $AHE$ is the orthocenter of $ABC$ but dilated towards $A$ with a factor of 1/2. Therefore, $$w=\frac{1}{2}((a+b+c)-a)+a=a+\frac{1}{2}(b+c).$$Now, $M=\frac{1}{2}(b+c)$, so we have $$w=a+m,$$therefore $OMWA$ is a parallelogram and the claim follows. Similarly, we have $CY=OM$ and parallel, so $WA\parallel CY$ and they are both the same length, so $WACY$ is a parallelogram. Similarly, $BXZD$ is a parallelogram. This means that the diagonals of $WXYZ$ are equal in length and parallel to the corresponding diagonals in $ABCD$, so the angle formed by the diagonals of $WXYZ$, which we call $\theta$, is the same as the diagonal angle of $ABCD$, so we have $$[WXYZ]=WY\cdot XZ\cdot \sin(\theta)=AC\cdot BD\cdot \sin(\theta)=[ABCD],$$so we are done.

Let $M$ be the orthocenter of $ABD$. Notice that if we take a homothety from $ABD$ to $AHE$ around $A$, we find that $W$ is the midpoint of $AM$, or $a+\frac{b+d}{2}$. Similarly, we find that $x=b+\frac{a+c}{2}$, $y=c+\frac{b+d}{2}$, and $z=d+\frac{a+c}{2}$. We find that $WY||AC$**, and $XZ||BD$**, and since $WY=AC$** and $XZ=BD$**, by the sine formula for area of a quadrilateral (which is just that the area is the diagonals multiplied, then multiplied by the angle between the two diagonals), we have that $[ABCD]=[WXYZ]$, and we are done. **All can be proved from the coordinates of $W$, $X$, $Y$, $Z$

$EFGH$ is a parallelogram because $EF\parallel AC\parallel GH$ and $EF=\frac{1}{2}AC=GH$. Given fixed $\omega$ and $B,C\in \omega$, if $A$ is a variable point on $\omega$ and $H$ is the orthocentre of $\triangle ABC$, we compute $AH=\lvert \tan\angle BAC\rvert BC$ which is fixed. Since $\angle EAH=\pi-\angle FCG$ and $EF=GH$, we have that $AW=CY$ and $AW, CY\perp BD\implies AW\parallel CY$. Thus, $AWYC$ is a parallelogram and similarly $BXZD$ is too. But now: $$\lvert ABCD\rvert =\frac{1}{2}AC\cdot BD\sin\angle(AC,BD)=\frac{1}{2} WY\cdot XZ\sin\angle(WY,XZ)=\lvert WXYZ\rvert$$as needed.

WLOG let us put $ABCD$ on complex unit circle. We define the midpoints as $e=\frac{a+b}{2}$, $f=\frac{b+c}{2}$, $g=\frac{c+d}{2}$, and $h=\frac{d+a}{2}$. Because $\triangle ABD$ is $\triangle AEH$ scaled down by 1/2, we can use $\triangle ABD$'s orthocenter and transform it to fit $\triangle AHE$. Since $\triangle ABD$ has orthocenter $a+b+d$, $\triangle AEH$ would have orthocenter $w= a+\frac{b+d}{2}$. Similarly we define $x = b+\frac{a+c}{2}$, $y=c+\frac{b+d}{2}$, and $z=d+\frac{c+a}{2}$. We want to calculate $[ABCD] = [ABC] + [ACD] = \dfrac{i}{4} \begin{vmatrix}a&\overline{a}&1\\ b&\overline{b}&1\\c&\overline{c}&1\end{vmatrix}+\dfrac{i}{4}\begin{vmatrix}a&\overline{a}&1\\ c&\overline{c}&1\\d&\overline{d}&1\end{vmatrix} = \dfrac{i}{4} \left( \dfrac{a}{b}+\dfrac{c}{a}+\dfrac{b}{c}-\dfrac{c}{b}-\dfrac{a}{c}-\dfrac{b}{a} \right )$. Now we wish to calculate $[WXYZ]$. $[WXYZ] = [WXY]+[WYZ] = \dfrac{i}{4} \begin{vmatrix}w&\overline{w}&1\\ x&\overline{x}&1\\y&\overline{y}&1\end{vmatrix}+\dfrac{i}{4}\begin{vmatrix}w&\overline{w}&1\\ y&\overline{y}&1\\z&\overline{z}&1\end{vmatrix}$. Which after a tedious bash leads us to an equivalent equation $\left( \dfrac{a}{b}+\dfrac{c}{a}+\dfrac{b}{c}-\dfrac{c}{b}-\dfrac{a}{c}-\dfrac{b}{a} \right )$. And we are done.

WLOG let ABCD be on the unit circle in the complex plane We know that $e = \frac{a+b}{2}$, $f = \frac{b+c}{2}$, $g = \frac{c+d}{2}$, and $h = \frac{a+d}{2}$. Thus, we can see that $w = 2a + \frac{b+d}{2}$, $x = 2b + \frac{a+c}{2}$, $y = 2c + \frac{b+d}{2}$, and $z = 2d + \frac{a+c}{2}$. From the complex shoelace formula (splitting the given into two triangles) we find that it remains to prove that $$[ABCD] = \frac{i}{4}\left( \begin{vmatrix} a & \overline{a} & 1 \\ b & \overline{b} & 1 \\ c & \overline{c} & 1 \end{vmatrix} + \begin{vmatrix} d & \overline{d} & 1 \\ b & \overline{b} & 1 \\ c & \overline{c} & 1 \end{vmatrix} \right) = \frac{i}{4}\left( \begin{vmatrix} a + \frac{b+d}{2} & \overline{a + \frac{b+d}{2}} & 1 \\ b + \frac{a+c}{2} & \overline{b + \frac{a+c}{2}} & 1 \\ c + \frac{b+d}{2} & \overline{c + \frac{b+d}{2}} & 1 \end{vmatrix} + \begin{vmatrix} d + \frac{a+c}{2} & \overline{d + \frac{a+c}{2}} & 1 \\ b + \frac{a+c}{2} & \overline{b + \frac{a+c}{2}} & 1 \\ c + \frac{b+d}{2} & \overline{c + \frac{b+d}{2}} & 1 \end{vmatrix} \right) = [WXYZ]$$ We can divide both sides by $\frac{i}{4}$ for simplicity. We can evaluate that $$\begin{vmatrix} a & \overline{a} & 1 \\ b & \overline{b} & 1 \\ c & \overline{c} & 1 \end{vmatrix} + \begin{vmatrix} d & \overline{d} & 1 \\ b & \overline{b} & 1 \\ c & \overline{c} & 1 \end{vmatrix} = a\overline{b} + \overline{a}c + b\overline{c} +d\overline{b} +\overline{d}c + b\overline{c} - \overline{a}b - \overline{b}c - a\overline{c} - \overline{d}b - \overline{b}c - d\overline{c} = \frac{a}{b} + \frac{a}{c} + \frac{b}{c} + \frac{d}{b} + \frac{c}{d} + \frac{b}{c} - \frac{b}{a} - \frac{c}{b} - \frac{a}{c} - \frac{b}{d} - \frac{c}{b} - \frac{d}{c}$$ Evaluating the other determinant we claim the the multiplication of the last fractional parts when added is equal to $0$. Looking at this multiplication we get $$\frac{b+d}{2}\overline{\frac{a+c}{2}} +\overline{\frac{b+d}{2}}\frac{b+d}{2} + \frac{a+c}{2}\overline{\frac{b+d}{2}} + \frac{a+c}{2}\overline{\frac{a+c}{2}} + \overline{\frac{a+c}{2}}\frac{b+d}{2} + \frac{a+c}{2}\overline{\frac{b+d}{2}}$$ The part we subtract would be $$\frac{b+d}{2}\overline{\frac{a+c}{2}} +\overline{\frac{b+d}{2}}\frac{b+d}{2} + \frac{a+c}{2}\overline{\frac{b+d}{2}} + \frac{a+c}{2}\overline{\frac{a+c}{2}} + \overline{\frac{a+c}{2}}\frac{b+d}{2} + \frac{a+c}{2} \overline{\frac{b+d}{2}}$$Thus we can see that they are equal. Now multiplying out all of the other terms (we group them by $a\overline{b}$ and $\overline{a}b$. Also notice that the $a\overline{a}$ terms cancel out within each of these groups) we get $$a \overline{b} - \overline{a}b + \frac{a \overline{c} - \overline{a}c + \overline{b}d -b \overline{d}}{2} + \overline{a}c - a \overline{c} + \frac{\overline{a}b + \overline{a}d +c \overline{b} + c \overline{d} - a \overline{b} - a \overline{d} - \overline{c}b -\overline{c}d}{2} + b\overline{c} - \overline{b}c + \frac{a\overline{c} +b\overline{d} - \overline{a}c - \overline{b}d}{2}$$This is for the first determinant. For the second determinant we get. $$d \overline{b} - \overline{d}b + \frac{a\overline{b} + c \overline{b} + \overline{a}d + \overline{c}d - \overline{a}b - \overline{c}b - a\overline{d} - c\overline{d}}{2} + d\overline{c} - \overline{d}c + \frac{a\overline{c} + d\overline{b} - \overline{a}c - \overline{d}b}{2}+ b\overline{c} - \overline{b}c + \frac{a\overline{c} +b\overline{d} - \overline{a}c - \overline{b}d}{2}$$ Adding these together we find that their sum is $\frac{a}{b} + \frac{a}{c} + \frac{b}{c} + \frac{d}{b} + \frac{c}{d} + \frac{b}{c} - \frac{b}{a} - \frac{c}{b} - \frac{a}{c} - \frac{b}{d} - \frac{c}{b} - \frac{d}{c}$. So the result is obivous. $\blacksquare$ SPHS1234 wrote: mathleticguyyy wrote: I wish I was alive in 2014 You weren't?? it is well known that one is not alive until they solve a tst problem. Also am I the only one who actually carried out the determinant. I literally came up with a brainless solution.

Set $(ABC)$ as the unit circle in the complex plane. Now $w = (2a + b + d)/2$, as we have a homothety of factor $2$ at $A$, and so $w = (a + b + d - a)/2 + a$. Similarly, $x = (2b + a + c)/2; y = (2c + b + d)/2; z = (2d + a + c)/2$. Now \begin{align*} [ABCD] &= [ADC] + [CBA] \\ &= \frac{i}{4} \begin{vmatrix} a &\bar a & 1 \\ d &\bar d & 1 \\ c &\bar c & 1 \\ \end{vmatrix} + \frac{i}{4} \begin{vmatrix} c &\bar c & 1 \\ b &\bar b & 1 \\ a &\bar a & 1 \\ \end{vmatrix} \\ &= \frac{i}{4} [(\bar d - \bar b)(a - c) + (\bar a - \bar c)(b - d)] \\ &= \frac{i}{4} \left[\frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d} - \frac{a}{b} - \frac{b}{c} - \frac{c}{d} - \frac{d}{a} \right]. \end{align*}However, \begin{align*} [WXYZ] &= [WZY] + [WYX] \\ &= \frac{i}{4} \begin{vmatrix} w & \bar w & 1 \\ z & \bar z & 1 \\ y & \bar y & 1 \\ \end{vmatrix} + \frac{i}{4} \begin{vmatrix} w & \bar w & 1 \\ y & \bar y & 1 \\ x & \bar x & 1 \\ \end{vmatrix} \\ &= \frac{i}{4}[(\bar z - \bar x)(w - y) + (\bar w - \bar y)(x - z)] \\ &= \frac{i}{4} \left[\frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d} - \frac{a}{b} - \frac{b}{c} - \frac{c}{d} - \frac{d}{a} \right], \end{align*}which is what we wanted to show. $\blacksquare$

Set $(ABCD)$ as the unit circle. Taking a homothety centered at $A$ with scale factor $2$, which takes $W$ to the orthocenter of $ABD$, yields that \[w = \frac{a + (a + b + d)}{2} = a + \frac{b + d}{2}\text{.}\]Clearly cyclic variations thereof also hold. Now note \[w - y = a + \frac{b + d}{2} - c - \frac{b + d}{2} = a - c\]so $WY \parallel AC$ and $WY = AC$. Similarly, $XZ \parallel BD$ and $XZ = BD$. Therefore, \begin{align*} |WXYZ| &= WY \cdot XZ \cdot \sin (\text{angle between }WY\text{ and }XZ) \\ &= AC \cdot BD \cdot \sin (\text{angle between }AC\text{ and }BD) \\ &= |ABCD| \end{align*}as desired.

How do you derive WY=AC without complex numbers? So many of the answers claim it's obvious but I can't seem to understand the synthetic/trig reasoning behind it. Any idea is appreciated.