AoPS - Problem

orl

22.10.2004 17:20

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

grobber

23.10.2004 11:50

Assume first that $ABCD$ is cyclic. It's easy then to see that $\angle APB+\angle CPD=\pi$, so $2S[APB]=R^2\sin \angle APB=R^2\sin\angle CPD=2S[CPD]$, where $R=PA=PB=PC=PD$ is the radius of the circumcircle of $ABCD$. Conversely, assume we have $S[APB]=S[CPD]$. Let $M,N$ be the midpoints of $AB,CD$ respectively. We have $TM=AM,TN=DN$, where $T=AC\cap BD$. The hypothesis is equivalent to $PM\cdot AM=PN\cdot DN$, and from here we deduce $PM\cdot TM=PN\cdot TN$. A quick angle chase reveals $\angle MPN=\angle MTN$, so the triangles $TMN,PNM$ are, in fact, congruent, meaning that $TMPN$ is a parallelogram. This, however, means that the triangles $TAM,DTN$ are similar (because they have three pairs of perpendicular sides), so $\angle TAB=\angle TDC$, Q.E.D.

jpark

22.01.2005 06:52

Problem 1. In the convex quadrilateral ABCD, the diagonals AC and BD are perpendicular and the opposite sides AB and DC are not parallel. The point P, where the perpendicular bisectors of AB and DC meet, is inside ABCD. Prove that ABCD is cyclic if and only if the triangles ABP and CDP have equal areas. I can prove this using an elegant method (I think) when ABCD is cyclic. I think there should be a nice way for this for the other way, but I can't find it.. Here is my half-solution. If someone could do the other half using the similar approach, it will be appreciated.. If ABCD is cyclic, then area ABP = area CDP Since the diagonals are perpendicular, 2*area ABCD = AB.CD + AD. BC (Ptolemy) Now, take the sectors ABP, BPC, CPD, DPA, and rearrange them, so that the side AB and CD would be adjacent to each other. It is still cyclic, because P is clearly the circumcentre of the circle. Now, the area of the new quadrilateral did not change, since we have only rearranged it. Let x be the angle between AB and CD. 2*area of new quadrilateral = AB.CD sin x + AD. BC sin (180-x) = sinx(AB.CD + AD. BC) Equate the two expressions, and we have sinx = 1, so x = pi/2. Simple calculation shows that angle APB = 180 - angle CPD, so the areas of the two triangles are equal. I've been trying to apply something similar to the other half, but have not had any progress so far.. Sorry for not using LaTeX.

yetti

22.01.2005 12:14

jpark wrote: Problem 1. In the convex quadrilateral ABCD, the diagonals AC and BD are perpendicular and the opposite sides AB and DC are not parallel. The point P, where the perpendicular bisectors of AB and DC meet, is inside ABCD. Prove that ABCD is cyclic if and only if the triangles ABP and CDP have equal areas. I can prove this using an elegant method (I think) when ABCD is cyclic. [...] I've been trying to apply something similar to the other half, but have not had any progress so far.. First, the circumcenter $O$ of a cyclic quadrilateral $ABCD$ with perpendicular diagonals is always in its interior. This is because the angle $\measuredangle ACB > \measuredangle AOB = 90^o$, hence, the chord $AB$ has the circumcenter on the same side as the vertices $C, D$. Similarly, all other sides of the quadrilateral. Let $ABCD$ be a quadrilateral with perpendicular diagonals, which is not cyclic. Let $M, N$ be the midpoints of the sides $AB, CD$ and $P$ the intersection of the perpendicular bisectors of these sides, which happens to be in the quadrilateral interior. Assume that the areas $S(\triangle APB), S(\triangle CPD)$ of the triangles $\triangle APB$ and $\triangle CPD$ are equal. Let $(O)$ be the circumcircle of the triangle $\triangle ABD$ centered at the point $O$ on the perpendicular bisector of $AB$. Assume first, that the vertex $C$ of the quadrilateral $ABCD$ is outside of this circumcircle. Move the point $C'$ along the line $AC$ form the point $C$ toward the the intersection $C\"$ of this line with the circle $(O)$. Let $N'$ be the midpoint of the segment $C'D$ and $P'$ the intersection of the perpendicular bisectors of $AB$ and $C'D$. In this process, the point $P'$ moves on the line $PM$ (the perpendicular bisector of $AB$) into the circumcenter $O$. The base $AB$ of the isosceles triangle $\triangle AP'B$ remains unchanged, but its altitude $P'M$ is increasing. As a result, the area $S(\triangle AP'B)$ is increasing and $S(\triangle AP'B) > S(\triangle APB)$. On the other hand, the base $C'D$ and the the angles $\measuredangle P'C'D = \measuredangle P'DC'$ of the isosceles triangle $\triangle C'P'D$ are decreasing. Hence, the remaining angle $\measuredangle C'P'D$ of this triangle is increasing and its altitude $P'N'$ is decreasing. As a result, the area $S(\triangle C'P'D)$ is decreasing and $S(\triangle C'P'D) < S(\triangle CPD)$. When and the point $C'$ coincides with the point $C\"$ and the point $P'$ coincides with the circumcenter $O$, the quadrilateral $ABC\"D$ becomes cyclic and $S(\triangle AOB) = S(\triangle C\"OD)$, as you just proved. But since $S(\triangle AOB) > S(\triangle APB)$, $S(\triangle C\"OD) < S(\triangle CPD)$, this is a contradiction with the assumption $S(\triangle APB) = S(\triangle CPD)$. The case, when the vertex $C$ lies inside of the circumcircle $(O)$ of the triangle $\triangle ABD$ is handled similarly.

Vo Duc Dien

25.11.2010 20:00

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.IMO1998Problem1 Vo Duc Dien

andrejilievski

10.09.2013 00:55

Let us assume that $ ABCD $ is cyclic, $ N,S $ be the midpoints of $ AB $ and $ DC $ respectively, $ AC $ and $ BD $ meet at $ R $, $ SR $ cuts $ AB $ at $ M $ and $ NR $ cuts $ CD $ at $ T $. According to Brahmagupta theorem $ SM \perp AB $ and $ TN \perp CD $ so $ SRNP $ is a parallelogram. Now, we have to show that $ PN*BN=PS*SC $ , but note that $ PN=RS=CS $ (since $ DRC $ is right angled and $ RS $ is its median ) and for the same reasons $ BN=RN=PS $ so we are done

DaChickenInc

27.06.2014 00:42

I tried to length-chase but it turned out to be a lot easier.

It seems easier to start with $ABCD$ being cyclic. Usually in these problems we can just backtrack or just use a linear function. Also, ROB1 noticed that since the perpendicular bisectors of $AB$ and $CD$ were diameters, so they met at the center, which means $P$ is the center of the circle circumscribing $ABCD$. When I tried to length-chase, I started by calling $Q$ the intersection of $AC$ and $BD$. By power of a point, if $a=AQ,b=BQ,c=CQ$, then $DQ=\frac{ac}{b}$. Then by Pythagorean theorem, $AB=\sqrt{a^2+b^2}$ and $CD=\sqrt{\frac{a^2c^2}{b^2}+c^2}$. If the desired areas are equal, then the ratios of the heights from $P$ would be $\frac{b}{c}$ in some order. So the heights must be nice in terms of $a,b,c$, which means the solution might not be far away. However, we only have one thing that gives us the height by Pythagorean theorem. That is, the distance from $A$ to the midpoint of $AB$ is $\frac{\sqrt{a^2+b^2}}{2}$, and the radius $AP$ is uniquely determined by $a,b,c$. But I was at a restaurant (Duke TIP rag night) and obviously too lazy to remember the circumradius formula, so I decided to switch to angle-chasing when length-chasing was straightforward. Well first of all, for properties of $ABP$ and $CDP$, we know that $AP=BP=CP=DP$ because they're radii. In what ways can we compare areas? That was when I suddenly remembered that if we proved $\angle APB+\angle CPD=180^\circ$, then we would be done. I fruitlessly angle-chased on the individual triangles, but then when I looked at the big picture, the circle, inscribed angles would work. $\angle APB+\angle CPD=2\angle ACB+2\angle CBD=180^\circ$, but we still have to prove in the other direction. Looking back at the diagrams I drew without the cyclic condition, I noticed that when you vary $D$, the point $P$ varies. We can only increase the area of one of the triangles and decrease the area of the other. If $ABCD$ is not cyclic, we have two similar cases. Assume $D$ lies outside the circumcircle of $ABC$. Let the perpendicular from $B$ to $AC$ meet the circumcircle of $ABC$ at $E$ and call $O$ the circumcenter of $ABC$. I tried this for a while but the variables weren't working in my favor, so I just approached it the other way. Let's use the right angles. Notice that if $M,N$ are the midpoints of $AB,CD$ respectively, and $Q$ was the intersection of $AC$ and $BD$, then $MB=MQ$ and $NC=NQ$. Since I stick to the simple methods, I will just use the simple base and height. The equality of areas means $MP\cdot MQ=MP\cdot MB=NP\cdot NC=NP\cdot NQ$. When only four points are involved in such an equation, we look for properties of quadrilateral $MPNQ$. Specifically, it looks like a parallelogram. We want to prove it using the products we already know, which call for similar triangles. It seems like we should prove that $\triangle MQN\sim\triangle NPM$, which is equivalent to $\angle MPN=\angle MQN$. Out of $\angle MPN$ and $\angle MQN$, it seems easiest based on the diagram to break up $\angle MQN$. We know that $\angle MQN=\angle MQA+90+\angle NQD$. It's a good thing we did not define new points here because that lets us notice $\angle MQA=\angle MAQ=\angle BAQ$, and $\angle NQD=\angle CDQ$. Then $\angle MQN=\angle BAQ+90+\angle CDQ$. Notice that $\angle BAQ=\angle BAC$ and $\angle CDQ=\angle CDB$. Here are some stupid things I did. Let's get rid of the $90$, so simplify $\angle BAQ+90$ first. We notice a compatible $90^\circ$ angle as $\angle AQB$, so $\angle BAQ+90=\angle BAQ+\angle AQB$. This reminds us of exterior angles. Which exterior angle is compatible with $\angle CDB$? Define $R$ as the intersection of $AB$ and $CD$. Then $\angle MQN=\angle BAQ+\angle AQB+\angle CDQ=\angle QBR+\angle CDQ=\angle DBR+\angle RDB=180-\angle BRD$. How do you compare this with $\angle MPN$? $\angle BRD=\angle MRN$, but $\angle PMR=\angle PNR=90^\circ$, so $\angle MQN=180-\angle MRN=\angle MPN$, as desired. This tells us that $MPNQ$ is a parallelogram. But we didn't know that $\angle PMQ=\angle PNQ$ before, so we will start from there. Clearly $90-\angle BMQ=\angle PMQ=\angle PNQ=90-\angle CNQ$, so $\angle BMQ=\angle CNQ$. Then we would uniquely define $\angle ABD$ and $\angle ACD$. But there is probably a shortcut.

Assume that $ABCD$ is cyclic. The perpendicular bisector of $AB$ is a line, and it contains the circumcenter of $ABCD$, so it is a diameter. The perpendicular bisector of $CD$ is also a diameter, and the diameters both contain the circumcenter, so the point they meet at, $P$, is the circumcenter of $ABCD$. Call $Q$ the intersection of $AC$ and $BD$. The problem tells us $\angle BQC=90^\circ$. By inscribed angles, $\angle APB+\angle CPD=2\angle ACB+2\angle CBD=2(\angle QCB+\angle CBQ)=2(180-\angle BQC)=180^\circ$. Rotate around $P$ by $\angle BPC$ counter-clockwise so that $C,D,P$ map to $B,D',P$ respectively. Then $\angle APB+\angle BPD'=\angle APB+\angle CPD=180^\circ$, which means $A,P,D'$ are collinear. Then the altitudes from $B$ to $AP$ and $D'P$ are equal. Also, $AP=D'P$, so the areas of $\triangle ABP$ and $\triangle BD'P$ are equal. But by rotation the area of $\triangle BD'P$ is equal to the area of $\triangle CDP$, so by transitivity we get the desired result. Call $M,N$ the midpoints of $AB,CD$ respectively. Since $\angle AQB=90^\circ$, $MB=MQ$, and similarly $NC=NQ$. If the areas of $\triangle ABP$ and $\triangle CDP$ were equal, then $MQ\cdot MP=MB\cdot MP=\frac{AB\cdot MP}{2}=\frac{CD\cdot NP}{2}=NC\cdot NP=NQ\cdot NP$. Partition $\angle MQN=\angle MQA+\angle AQD+\angle DQN$. Notice that $\angle AQD=90^\circ=\angle AQB$, $\angle MQA=\angle MAQ=\angle BAQ$, and similarly $\angle DQN=\angle CDQ$, so $\angle MQN=\angle AQB+\angle BAQ+\angle CDQ$. Define $R$ to be the intersection of $AB$ with $CD$. Then by exterior angles, $\angle MQN=\angle QBR+\angle CDQ=\angle DBR+\angle RDB$. Since the sum of angles in a triangle is $180^\circ$, $\angle MQN=180-\angle BRD=180-\angle MRN$. Notice that $360=\angle MRN+\angle RNP+\angle NPM+\angle PMR=180+\angle MRN+\angle MPN$. Subtracting $180+\angle MRN$ from both sides gives us $\angle MPN=180-\angle MRN=\angle MQN$. Recall from earlier that $MQ\cdot MP=NQ\cdot NP$. This means $\frac{MQ}{NP}=\frac{NQ}{MP}$. By our new result that $\angle MPN=\angle MQN$, SAS tells us $\triangle MQN\sim\triangle NPM$. Then $\angle PMQ=\angle PMN+\angle NMQ=\angle QNM+\angle MNP=\angle PNQ$. Clearly $\angle MBQ=\angle MQB$, so by exterior angles, $\angle AMQ=\angle MBQ+\angle MQB=2\angle MBQ=2\angle ABD$. Analogously, $\angle DNQ=2\angle ACD$. But $\angle ABD=\frac{\angle AMQ}{2}=\frac{90+\angle PMQ}{2}=\frac{90+\angle PNQ}{2}=\frac{\angle DNQ}{2}=\angle ACD$, so $ABCD$ is cyclic.

WJ.JamshiD

21.05.2015 18:04

Which is Brahmagupta theorem?

anantmudgal09

01.01.2020 00:43

Storage 1998 G1 wrote: A convex quadrilateral $ABCD$ has perpendicular diagonals. The perpendicular bisectors of the sides $AB$ and $CD$ meet at a unique point $P$ inside $ABCD$. Prove that the quadrilateral $ABCD$ is cyclic if and only if triangles $ABP$ and $CDP$ have equal areas. If $ABCD$ cyclic with circumradius $R$, then conclusion follows since $\angle APB+\angle CPD=180^{\circ}$ as $AC \perp BD$, so $[APB]=\frac{1}{2}R^2\sin \angle APB=\frac{1}{2}R^2\sin \angle CPD=[CPD]$ as desired. Now assume $ABCD$ is not cyclic, but $AC \perp BD$ and $[APB]=[CPD]$. Let $E=AC \cap BD$. WLOG, say $P$ lies inside $\triangle BEC$ and $PA<PD$. Let $A', B'$ be second intersections of diagonals $AC$ and $BD$ with $\odot(P, PA)$. Note that $ABA'B'$ is a convex cyclic quadrilateral with perpendicular diagonals, and so $[APB]=[A'PB']$. Note that $A'$ lies on segment $AC$ and $B'$ on segment $BD$, since $PA<PD$. Since ray $\overrightarrow{A'P}$ meets line $BD$ at it's extension beyond $P$ (as $\angle EA'P=\angle PAC<90^{\circ}$) and $B', D$ lie on the same side of line $A'P$, with $B'$ closer to the intersection than $D$, we conclude that $[A'PB']<[A'PD]$. But $[A'PD]<[BPD]$ since $A'$ lies inside $\triangle PBD$. This chain of inequalities is a contradiction! So, we conclude that $ABCD$ must be cyclic. $\blacksquare$ P.S. grobber's solution is really nice

v_Enhance

04.03.2021 18:01

Solution from Twitch Solves ISL: If $ABCD$ is cyclic, then $P$ is the circumcenter, and $\angle APB + \angle PCD = 180^{\circ}$. The hard part is the converse. [asy][asy] import graph; size(10cm); pen zzttqq = rgb(0.6,0.2,0.); pen aqaqaq = rgb(0.62745,0.62745,0.62745); pen cqcqcq = rgb(0.75294,0.75294,0.75294); draw((2.28154,-9.32038)--(8.62921,-11.79133)--(2.93422,-2.54008)--cycle, linewidth(0.6) + zzttqq); draw((2.28154,-9.32038)--(-2.,-4.)--(-4.,-12.)--cycle, linewidth(0.6) + zzttqq); draw((8.62921,-11.79133)--(-2.,-4.), linewidth(0.6)); draw((2.93422,-2.54008)--(-4.,-12.), linewidth(0.6)); draw((8.62921,-11.79133)--(2.93422,-2.54008), linewidth(0.6)); draw((-2.,-4.)--(-4.,-12.), linewidth(0.6)); draw((2.28154,-9.32038)--(5.78172,-7.16571), linewidth(0.6) + blue); draw((2.28154,-9.32038)--(-3.,-8.), linewidth(0.6) + blue); draw((2.28154,-9.32038)--(8.62921,-11.79133), linewidth(0.6) + zzttqq); draw((8.62921,-11.79133)--(2.93422,-2.54008), linewidth(0.6) + zzttqq); draw((2.93422,-2.54008)--(2.28154,-9.32038), linewidth(0.6) + zzttqq); draw((2.28154,-9.32038)--(-2.,-4.), linewidth(0.6) + zzttqq); draw((-2.,-4.)--(-4.,-12.), linewidth(0.6) + zzttqq); draw((-4.,-12.)--(2.28154,-9.32038), linewidth(0.6) + zzttqq); draw((2.93422,-2.54008)--(-0.31533,2.73866), linewidth(0.6)); draw((-2.,-4.)--(-0.31533,2.73866), linewidth(0.6)); draw(circle((2.25532,-9.31383), 6.80768), linewidth(0.6) + aqaqaq); draw((0.51354,-5.84245)--(-3.,-8.), linewidth(0.6) + blue); draw((0.51354,-5.84245)--(5.78172,-7.16571), linewidth(0.6) + blue); dot("$D$", (-4.,-12.), dir((-76.113, -27.810))); dot("$C$", (-2.,-4.), dir((-75.616, 41.734))); dot("$B$", (2.93422,-2.54008), dir((1.940, 37.397))); dot("$A$", (8.62921,-11.79133), dir((31.751, -33.422))); dot("$P$", (2.28154,-9.32038), dir((-11.247, -66.945))); dot("$M$", (5.78172,-7.16571), dir((39.728, 27.050))); dot("$N$", (-3.,-8.), dir((-82.402, 23.307))); dot("$E$", (0.51354,-5.84245), dir((-30.585, 62.531))); dot("$X$", (-0.31533,2.73866), dir((-23.973, 36.915))); [/asy][/asy] Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{CD}$. Claim: Unconditionally, we have $\measuredangle NEM = \measuredangle MPN$. Proof. Note that $\overline{EN}$ is the median of right triangle $\triangle ECD$, and similarly for $\overline{EM}$. Hence $\measuredangle NED = \measuredangle EDN = \measuredangle BDC$, while $\measuredangle AEM = \measuredangle ACB$. Since $\measuredangle DEA = 90^{\circ}$, by looking at quadrilateral $XDEA$ where $X = \overline{CD} \cap \overline{AB}$, we derive that $\measuredangle NED + \measuredangle AEM + \measuredangle DXA = 90^{\circ}$, so \[ \measuredangle NEM = \measuredangle NED + \measuredangle AEM + 90^{\circ} = -\measuredangle DXA = -\measuredangle NXM = -\measuredangle NPM \]as needed. $\blacksquare$ However, the area condition in the problem tells us \[ \frac{EN}{EM} = \frac{CN}{CM} = \frac{PM}{PN}. \]Finally, we have $\angle MEN > 90^{\circ}$ from the configuration. These properties uniquely determine the point $E$: it is the reflection of $P$ across line $MN$. So $EMPN$ is a parallelogram, and thus $\overline{ME} \perp \overline{CD}$. This implies $\measuredangle BAE = \measuredangle CEM = \measuredangle EDC$ giving $ABCD$ cyclic.

bluelinfish

19.07.2021 17:13

Way too hard for an early IMO #1. First, we will prove that if $ABCD$ is cyclic, the area condition holds. Let $E$ be the intersection of lines $AC$ and $BD$, and let $r$ be the circumradius of $ABCD$. Note that $P$ is the circumcenter of $ABCD$, so reflecting $C$ about $P$ to a point $C'$ creates a trapezoid $AC'BD$. In particular, because it is cyclic, it is an isosceles trapezoid, so $\angle CPB=180^{\circ}-\angle BPC'=180^{\circ}-\angle APD$. Therefore $\angle APB+\angle CPD=180^{\circ}$, so $\sin \angle APB =\sin \angle CPD$, implying the area condition as the area of $APB$ and $CPD$ are $\frac{r^2}{2} \sin \angle APB$ and $\frac{r^2}{2} \sin \angle CPD$, respectively. Now we will prove the harder part, that given the area of triangles $APB$ and $CPD$ are the same, $ABCD$ is cyclic. Let $X$ be the intersection of lines $AB$ and $CD$ (if $AB\parallel CD$, either $P$ doesn't exist or $ABCD$ is an isosceles trapezoid), and let $M$ and $N$ be the midpoints of $AB,CD$, respectively. Then notice that \begin{align*} \angle MPN &= 180^{\circ}-\angle BXC \\ &=\angle XBC+\angle XCB \\ &= \angle ABE+(\angle EBC+\angle ECB)+\angle ECD \\ &= \angle BME+90^{\circ}+\angle NEC \\ &= \angle MEN. \end{align*}Also, we have $$\frac{EN}{EM}=\frac{CN}{BM}=\frac{PM}{PN},$$where the second equality is by the area condition. This implies $EMPN$ is a parallelogram. Therefore we have $$\angle ACD = \frac{1}{2}\angle END = \frac{1}{2}(90^{\circ}-\angle PNE) = \frac{1}{2}(90^{\circ}-\angle PME) =\frac{1}{2}\angle AME =\angle ABD,$$and we are done.

Attachments:

asdf334

01.01.2022 03:26

I don't know if this works (there are probably a lot of configuration issues) but I think you can fix $A,B,C$ and let $D$ vary along the line through $B$ that is perpendicular to $AC$ and show that only one point on this line will work (by showing that the area of one triangle decreases while the other increases), and then check that this point must be the intersection of $(ABC)$ with the line. Edit: I don't think there would be any issues with area becoming "negative" because we are given $P$ is inside $ABCD$

Rekt4

17.01.2022 22:01

Let $E = \overline{AC} \cap \overline{BD}$. Suppose $P$ lies in $\triangle AEB$. The other cases are similar to the following. Let $M,N$ be the foot of the perpendiculars from $P$ to $AC$ and $BD$ respectively. We have \begin{align*} [PAB] &= \dfrac 12 (AE\cdot BE - BE\cdot ME - AE\cdot EN) \\ &= \dfrac 12 (AM\cdot BN - ME\cdot NE) \end{align*}Similarly, one can see that $[PCD] = \frac 12 (CM\cdot ND - EM\cdot EN)$. Thus, we have \begin{align} [PAB] - [PCD] = \dfrac 12 (AM\cdot BN - CM\cdot DN) \end{align}Suppose $ABCD$ is cyclic, then $P$ is the circumcenter of $ABCD$. So, $AM = CM$ and $BN = DN$. Thus, $(1)$ gives us $[PAB] = [PCD]$. On the other hand, suppose $ABCD$ is not cyclic. Without the loss of generality assume $PA = PB > PC = PD$. Then we see that $AM > CM$ and $BN > DN$, so $(1)$ gives us $[PAB] > [PCD]$. This proves the other implication, and we are done.

Ruy

17.01.2022 22:24

orl wrote: A convex quadrilateral $ABCD$ has perpendicular diagonals. The perpendicular bisectors of the sides $AB$ and $CD$ meet at a unique point $P$ inside $ABCD$. Prove that the quadrilateral $ABCD$ is cyclic if and only if triangles $ABP$ and $CDP$ have equal areas. I guess you can successfully complex-bash this using the axis as diagonals with the standard aereal formula and the power of the origin point.

Davsch

15.02.2022 00:22

Ruy wrote: I guess you can successfully complex-bash this using the axis as diagonals with the standard aereal formula and the power of the origin point. Well, even cartesian coordinates are enough. Let $A=(0,a),B=(b,0),C=(0,c),D=(d,0)$. $ABCD$ is cyclic if and only if $ac=bd$. By perpendiculars from $\left(\frac b2,\frac a2\right)$ and $\left(\frac d2,\frac c2\right)$, $P=\left(\frac{b^2c-a^2c-ad^2+ac^2}{2(bc-ad)},\frac{bc^2+b^2d-bd^2-a^2d}{2(bc-ad)}\right)$. Using determinants, the equation $[ABP]=[CDP]$ factors as $0=(bd-ac)((b-d)^2+(a-c)^2)$, as desired.

Mahdi_Mashayekhi

23.03.2022 08:31

Let diagonals meet at $S$ and Let $M,N$ be midpoint's of $AB,CD$. we have $[APB] = [CPD] \implies [AMP] = [DNP] \implies \frac{MS}{NS} = \frac{NP}{MP}$. we also have $\angle MSN = \angle MAS + \angle NDS + \angle 90 = \angle MPN$ which implies that $MPN$ and $NSM$ are congruent so $MPNS$ is parallelogram so $SM \perp DN$ and $SN \perp AM$ and $AS \perp DS$ which implies $AMS$ and $DNS$ are similar so $\angle MAS = \angle NDS$ so $ABCD$ is cyclic. Now assume $ABCD$ is cyclic. $2[APB] = AP.PB.\sin{APB} = DP.PC.\sin{DPC} = 2[DPC] \implies [APB] = [CPD]$

asdf334

10.07.2022 00:19

Let $A=(0,2a)$, $B=(2b,0)$, $C=(0,2c)$, $D=(2d,0)$, where $a,b>0$ and $c,d<0$. Then we have: \[AB:y=\frac{b}{a}x-\frac{b^2}{a}+a\]\[CD:y=\frac{d}{c}x-\frac{d^2}{c}+c\]and their intersection has $x$-coordinate $\frac{b^2c+ac^2-ad^2-a^2c}{bc-ad}.$ Clearly the length of the altitude from $P$ to $AB$ is then \[\left(b-\frac{b^2c+ac^2-ad^2-a^2c}{bc-ad}\right)\cdot \frac{\sqrt{a^2+b^2}}{a}\]while the length of the base, $AB$, is $2\sqrt{a^2+b^2}$. Then the area of $ABP$ is equal to \[\left(\frac{d^2+ac-c^2-bd}{bc-ad}\right)(a^2+b^2).\]Similarly, the length of the altitude from $P$ to $CD$ is equal to \[\left(\frac{b^2c+ac^2-ad^2-a^2c}{bc-ad}-d\right)\cdot \frac{\sqrt{c^2+d^2}}{-c}\]and the length of the base is $2\sqrt{c^2+d^2}$. Then the area of $CDP$ is equal to \[-\left(\frac{b^2+ac-a^2-bd}{bc-ad}\right)(c^2+d^2).\]If $ABP$ and $CDP$ have equal areas then \[(d^2+ac-c^2-bd)(a^2+b^2)=-(b^2+ac-a^2-bd)(c^2+d^2)\]which is obviously true when $ac=bd$, which is precisely the condition that $ABCD$ is cyclic. Then, it suffices to show that from this equation follows $ac=bd$; note that the equation is equivalent to \[(ac-bd)(a^2+b^2+c^2+d^2)=(a^2-b^2)(c^2+d^2)+(a^2+b^2)(c^2-d^2)=2a^2c^2-2b^2d^2.\]Then if $ac\neq bd$ then we have that $a^2+b^2+c^2+d^2=2ac+2bd$ which implies that $a=c$ and $b=d$, a contradiction because $a,b>0$ and $c,d<0$. We are done. $\blacksquare$

awesomeming327.

13.07.2022 03:51

Forward Direction If $ABCD$ is cyclic then $P$ must be the circumcenter. Since $AC\perp BD$ we have $\widehat{AB}+\widehat{CD}=180^\circ.$ Thus, $\angle APB,\angle CPD$ are supplementary. In particular, $\sin (\angle APB)=\sin (\angle CPD).$ Let $r$ be the circumradius then \[[APB]=\frac{1}{2}r^2\sin (\angle APB)=\frac{1}{2}r^2\sin (\angle CPD)=[CPD]\]as desired. Backward Direction If $[APB]=[CPD]$ then let $E$ be the intersection of $AC$ and $BD.$ Let $M$ be midpoint of $CD$ and $N$ be midpoint of $AB.$ We claim that $EMPN$ is a parallelogram. To prove our claim, let $E=(0,0),A=(0,a),B=(b,0),C=(0,-c),D=(-d,0)$ where $a,b,c,d$ are positive. $M=(-\frac{d}{2},-\frac{c}{2})$ and $N=(\frac{b}{2},\frac{a}{2}).$ It suffices to show that $P=(\frac{b-d}{2},\frac{a-c}{2}).$ Let $P'=(\frac{b-d}{2},\frac{a-c}{2})$ and we verify that $P'C=PD$ and $P'A=P'B.$ Since $P$ is unique, we know that $P'=P$ so our claim holds. Since $EMPN$ is a parallelogram, $EM=PN$, which implies $DM=PN.$ We have $[DMP]=[APN]$ which implies $\triangle DMP\cong \triangle PNA.$ Thus, $DP=AP$ and so $P$ is equidistant from $A,B,C,D.$ It follows that $ABCD$ is cyclic, as desired.

bobthegod78

16.09.2022 03:37

The direction from cyclic to equal area is pretty easy so I'll skip it. Here's a really easy way for equal area to cyclic (i don't think anyone's posted a coordinate method yet? or at least the easy way). Let $P=(0,0)$ and let the circle centered at $P$ through $AB$ be $x^2+y^2=1$ and through $C,D$ be $x^2+y^2=r^2.$ Now wlog the perpendicular lines to be perpendicular to the axes. We can wlog some more variables and find that $A(-\sqrt{1-y^2}, y), B(x, \sqrt{1-x^2}), C(\sqrt{r^2-y^2}, y), D(x, -\sqrt{r^2-x^2}).$ Now Shoelace trivializes the problem by giving us that $\sqrt{(r^2-x^2)(r^2-y^2)} = \sqrt{(1-x^2)(1-y^2)}$ and we get $r=1$, as desired.

Taco12

12.12.2022 01:34

Apply cartesian coordinates with $A=(0,a), C=(0,c), B=(b,0), D=(d,0)$ so that the diagonals lie on the axis of the coordinate plane. Then we have the perpendicular bisector of $AB$ as $y=\frac{b}{a}x+\frac{a}{2}-\frac{b^2}{2a}$ and similarly the perpendicular bisector of $CD$ is $y=\frac{d}{c}x+\frac{c}{2}-\frac{d^2}{2c}$. Then, we have $$P=\left(\frac{ac^2+b^2c-a^2c-ad^2}{2bc-2ad},\frac{bc^2+b^2d-bd^2-a^2d}{2bc-2ad}\right)$$and Shoelace now gives $$-a^3c + a^2 b d + 2 a^2 c^2 - a b^2 c - a c^3 - a c d^2 + b^3 d - 2 b^2 d^2 + b c^2 d + b d^3=0 \rightarrow ac=bd,$$which implies cyclicity. $\blacksquare$

awesomehuman

26.03.2023 23:42

Assume $ABCD$ is cyclic. Then $P$ is the center of $(ABCD)$. We have $\angle APB+\angle CPD=2\angle ACB+2\angle CAD=180$. So, $AP*BP*sin(APB)=CP*DP*sin(CPD)$. So, $[ABP]=[CDP]$. Assume $[ABP]=[CDP]$. Then, $[ABPC]=[CDPB]$. Let $X$ and $Y$ be the feet of the perpendiculars from $P$ to $AC$ and $BD$ respectively. Then, $[ABYC]=[CDXB]$. So, $CX/AX=DY/BY$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$ respectively. Then, $PY^2+BY^2=BP^2=AP^2=PX^2+AX^2$. Similarly, $PY^2+DY^2=PX^2+CX^2$. So, $CX^2-AX^2=DY^2-BY^2$. Combining this with $CX/AX=DY/BY$, we find that $DY=BY$ and $CX=AX$ (in which case $ABCD$ is cyclic as desired) or $AC=BD$. In this case, $PY=PX$, so by symmetry, $ABCD$ is an isosceles trapezoid, which is cyclic.

huashiliao2020

28.05.2023 01:03

Necessity: Given ABCD is cyclic it is well known that the intersection P is the unique circumcenter. <ADB+<CAD=90 degrees (orthogonal quad), so <APB+<CPD=180 degrees. Then (CP=DP=AP=BP) CP*DPsinCPD=AP*BPsinAPB, or [ABP]=[CBP]. (directed angles might be necessary depending on config? can someone pm or tell me here if it is necessary, the sketch is done though) Sufficiency: Given [ABP]=[CBP] I use coordinates. Unfortunately my proof was not saved but basically I let the intersection of the diagonals be the origin, and shoelace formula to get PoP stays constant on the origin with ac=bd. Actually, this can lead to an interesting corollary: BF=EP, from which it follows that EC=PF.

Attachments:

lpieleanu

14.07.2023 16:22

ezpotd

21.08.2024 05:31

WLOG let $AC$ be the $y$ axis, $BD$ be the $x$ axis, WLOG let $A,D$ have nonnegative coordinates, $B,C$, nonpositive . Then fix points $P,C,D$, WLOG let $P$ lie within $OCD$ where $O$ is the origin. Let $A = (0,a), B = (-b, 0 )$, with $a,b> 0$. Note that for each value of $a$, the point $B$ lies on the circle with radius $AP$ centered at $P$, and the $x$ axis, but this circle intersects the $x$ axis at two points, one of which is on the wrong side, so $b$ is a function of $a$. We show this function is increasing. Let $P = (x,-y)$ with $x,y > 0$, then $(b + x)^2 + y^2 = (a + y)^2 + x^2$, so clearly as $a$ increases, so does $(a + y)$ since $a, y > 0$, and so does $(a + y)^2 $ , so $(b + x)^2$ is forced to increase, since $b + x > 0$ we also see that $b + x$ is forced to increase, so $b$ is forced to increase. Now we show that this property implies that the area of $APB$ is increasing as a function of $a$. It suffices to show that the areas of $AOB, AOP, BOP$ increase, but this is trivial by sine area formula and observing that all side lengths are constant or increasing. Thus for each fixture of $C,P,D$, exactly one position of $A$ forces a position of $B$ such that $APB$ and $CPD$ have equal areas. It suffices to prove that if $ABCD$ is cyclic, then $APB$ and $CPD$ have equal areas. This is because we have shown $APB$ and $CPD$ are equal areas in exactly one position position of $A$, so it will be then implied that position is exactly the one for which $ABCD$ is cyclic, hence proving the only if direction. If $ABCD$ is cyclic, then we see it has circumcenter $P$, so we can calculate the area of $APB$ as $\frac 12 R^2 \sin 2 \angle ACB$, symmetrically the other area is $\frac 12 R^2 \sin 2 \angle CAD$, so it remains to prove $2 \angle ACB$ and $2 \angle CAD$ are supplements, which is obvious. It's obvious all possible quadrilaterals $ABCD$ are handled via this argument, but it should be noted.