IMO ShortList 1998, geometry problem 1

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Source: IMO ShortList 1998, geometry problem 1

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orl

#1 h Oct 22, 2004, 2:20 PM • 4 Y

Y by Bachsonata3, Adventure10, lian_the_noob12, Mango247

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.

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orl

#2 h Oct 22, 2004, 2:20 PM • 3 Y

Y by Bachsonata3, Adventure10, Mango247

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

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#3 h Oct 23, 2004, 8:50 AM • 9 Y

Y by numerology9, ArefS, anantmudgal09, Illuzion, Bachsonata3, Adventure10, Mango247, Deadline, and 1 other user

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.

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jpark

#4 h Jan 22, 2005, 4:52 AM • 3 Y

Y by Bachsonata3, Adventure10, Mango247

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.

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yetti

#5 h Jan 22, 2005, 10:14 AM • 3 Y

Y by Bachsonata3, Adventure10, Mango247

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.

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#6 h Nov 25, 2010, 6:00 PM • 3 Y

Y by Bachsonata3, Adventure10, Mango247

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

Vo Duc Dien

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#7 h Sep 9, 2013, 9:55 PM • 3 Y

Y by Bachsonata3, Adventure10, Mango247

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

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#8 h Jun 26, 2014, 9:42 PM • 3 Y

Y by Bachsonata3, Adventure10, Mango247

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

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#9 h May 21, 2015, 3:04 PM • 2 Y

Y by Adventure10, Mango247

Which is Brahmagupta theorem?

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#10 h Dec 31, 2019, 9:43 PM • 1 Y

Y by Adventure10

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

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#11 h Mar 4, 2021, 3:01 PM • 6 Y

Y by Pluto04, Aimingformygoal, nikitadas, HamstPan38825, ike.chen, khina

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.
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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.

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#12 h Jul 19, 2021, 2:13 PM • 1 Y

Y by centslordm

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.

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#13 h Jan 1, 2022, 12:26 AM

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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 $ABCD$

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Rekt4

#14 h Jan 17, 2022, 7:01 PM

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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.

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Ruy

#15 h Jan 17, 2022, 7:24 PM

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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.

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Davsch

#16 h Feb 14, 2022, 9:22 PM

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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.

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#17 h Mar 23, 2022, 5:31 AM

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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. $2[APB] = AP.PB.\sin{APB} = DP.PC.\sin{DPC} = 2[DPC] \implies [APB] = [CPD]$

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#18 h Jul 9, 2022, 9:19 PM • 1 Y

Y by Sross314

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. $\blacksquare$

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awesomeming327.

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awesomeming327.

#19 h Jul 13, 2022, 12:51 AM

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Diagram

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.

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#20 h Sep 16, 2022, 12:37 AM • 2 Y

Y by Mango247, Mango247

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.

This post has been edited 1 time. Last edited by bobthegod78, Sep 16, 2022, 12:37 AM

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Taco12

#21 h Dec 11, 2022, 10:34 PM

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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. $\blacksquare$

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#22 h Mar 26, 2023, 8:42 PM

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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.

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#26 h May 27, 2023, 10:03 PM

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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.

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#27 h Jul 14, 2023, 1:22 PM

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only if (forward direction)
if (backward direction)

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ezpotd

#28 h Aug 21, 2024, 2:31 AM

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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.

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#29 h Apr 13, 2025, 7:53 PM

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Bashed on paper; only summary provided.

The forwards direction is obvious (sine area formula) so we prove the backwards direction.

Use Cartesian coordinates with $A=(a,0)$ , $B=(0,b)$ , $C=(c,0)$ , $D=(0,d)$ , $P=(x,y)$ . Wait I just realized that I seem to have swapped $A$ and $C$ but of course that's not an issue.

Indeed, by the Shoelace formula, the second condition is equivalent to $(a-c)y=(b-d)x+(2da-2bc)$ . The perpendicular bisectors are $by=cx+b^2-c^2$ and $dy=ax+d^2-a^2$ . Then the second condition is equivalent to the following determinant vanishing:
$\begin{vmatrix} a-c & b-d & 2da-2bc \\ b & c & b^2-c^2 \\ d & a & d^2-a^2 \end{vmatrix}.$ After fully expanding then factoring, we eventually get that this determinant is $(bd-ac)((a-c)^2+(b-d)^2)$ , which can only vanish when $ac=bd$ , as desired. $\blacksquare$

This post has been edited 1 time. Last edited by cj13609517288, Apr 13, 2025, 7:56 PM

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