I spent 4 hours on this and got nothing...

This has a 5 minute complex solution by extending $AX,CX$ to hit the circle again.

Wow... okay. I tried using Trig Ceva, but it got too messy Hopefully 1 point, but probably not from the looks of it

Wait really? I had a long complex bash.

Yeah, you get $X\in BD\iff (a+c)(b+d)=2(ac+bd)$ and it stays the same under rotation of variables so yay

Did anyone trig bash this problem? I used law of sines on a bunch of triangles. Also, isn't the only if part similar to the if part? So could you say like if we proved the "only if" part, then the "if" part follows similarly?

I basically said that X exists if and only if $\frac{AB}{AD}=\frac{CB}{CD}$ and proved that, then Y exists if and only if $\frac{BA}{BC}=\frac{DA}{DC}$ which is just a rearrangement of the other condition, so they imply each other

infiniteturtle wrote: Yeah, you get $X\in BD\iff (a+c)(b+d)=2(ac+bd)$ and it stays the same under rotation of variables so yay Okay good I got that equation.

I showed that if the angle bisectors of A and C intersected at BD, then X exists. Then I showed that if the angle bisectors of A and C intersected at BD, then the angle bisectors of B and D intersect at AC, implying Y exists.

Is it true that AC and BD were perpendicular? I showed that, I'm hoping to get some points. I also stated that the statement and its converse were the same thing essentially, does that warrant points?

mathgenius64 wrote: Is it true that AC and BD were perpendicular? I showed that, I'm hoping to get some points. No. The condition is equivalent to $ABCD$ being a harmonic quadrilateral mathgenius64 wrote: I also stated that the statement and its converse were the same thing essentially, does that warrant points? No.

I proved that $X\ exists\Longleftrightarrow (AB)(CD)=(BC)(DA)$ and similarly for $Y$, does this work?

This man is a legend, he does it again: http://www.artofproblemsolving.com/community/c2595h1075117

v_Enhance wrote: mathgenius64 wrote: Is it true that AC and BD were perpendicular? I showed that, I'm hoping to get some points. No. The condition is equivalent to $ABCD$ being a harmonic quadrilateral mathgenius64 wrote: I also stated that the statement and its converse were the same thing essentially, does that warrant points? No. So do we really need to prove both if and only if parts? Could we at least say that one direction was similar to the other or not?

infiniteturtle wrote: This has a 5 minute complex solution by extending $AX,CX$ to hit the circle again. Darn I had this idea but had only a few minutes left when I did this... shoot. I angle chased and it turned out to work, since I used a lot of intercepted arcs, but ran out of time to write up...

Yet another JMO problem that immediately dies to projective...

Yeah, I was thinking about projective, but I couldn't remember Pascal's theorem. Got that X had to be midpoint of BD, but I couldn't turn that into a full proof. I'm guessing 2 points, 4 if I'm REALLY lucky.

It's just a well known symmedian lemma which leads to ABCD harmonic.

Let $AC \cap BD = Z$, and assume the existence of a satisfactory point $X$. Then Steiner's Ratio Theorem yields $$\left( \frac{AB}{AD} \right)^2 = \frac{BX}{XD} \cdot \frac{BZ}{ZD} = \left(\frac{CB}{CD} \right)^2$$so $ABCD$ is harmonic. Redefine $Y$ as the midpoint of $AC$. Isogonality and properties of harmonic quadrilaterals gives $\angle CBD = \angle YBA$ and $\angle CDB = \angle YDA$, i.e. the existence of a satisfactory point $X$ implies the existence of a satisfactory point $Y$. The desired result follows immediately from symmetry. $\blacksquare$ Remarks: This result is extremely well-known. I have no idea why this question made it onto the exam.

Can I please have feedback?

Why isn't this in the chapter 9 excercise of EGMO even when #3 was in 2 exercises tenniskidperson3 wrote: Let $ABCD$ be a cyclic quadrilateral. Prove that there exists a point $X$ on segment $\overline{BD}$ such that $\angle BAC=\angle XAD$ and $\angle BCA=\angle XCD$ if and only if there exists a point $Y$ on segment $\overline{AC}$ such that $\angle CBD=\angle YBA$ and $\angle CDB=\angle YDA$. By application Isogonal Conjugate ratios in $\triangle ABD$ and $\triangle BCD$, $$(\frac{AB}{AD})^2=\frac{BE}{DE}.\frac{BX}{DX}=(\frac{BC}{CD})^2 \implies \frac{AB}{AD}=\frac{BC}{CD} \implies \text{ABCD is harmonic!}$$Since $AX$ and $AC$ are isogonal and $AC$ is the $\text{A-Symmedian}$ by definition of harmonic quadrilateral,so $X$ is the midpoint of $BD$. Similarly taking $Y$ as the midpoint of $AC$ completes the proof !

Let $E=\overline{AC}\cap\overline{BD}.$ Suppose $X$ is on $\overline{BD}$ such that $\angle BAC=\angle XAD$ and $\angle BCA=\angle XCD.$ By Isogonal Ratios, $$\frac{AB^2}{AD^2}=\frac{BX}{DX}\cdot\frac{BE}{DE}\quad\text{and}\quad\frac{BC^2}{CD^2}=\frac{BX}{DX}\cdot\frac{BE}{DE}$$so $ABCD$ is harmonic. Hence, $X$ is the midpoint of $\overline{BD}$ and letting $Y$ be the midpoint of $\overline{AC}$ suffices. $\square$

wow how did i not see this before We claim that both conditions are equivalent to $ABCD$ being harmonic. Let $P$ be the intersection of the diagonals. Note that $AP$ and $AX$ are isogonal in $\triangle ABD$, and $CX$ and $CP$ are isogonal in $\triangle DBC.$ If $X$ exists, then $$\frac{DX}{BX}\cdot\frac{DP}{BP}=(\frac{AD}{AB})^2=(\frac{DC}{BC})^2\rightarrow AD\cdot BC=DC\cdot AB.$$ Similarly, if $Y$ exists implies that $ABCD$ is harmonic. Now, let's show the other direction. If $ABCD$ is harmonic, then let $X$ be such that $AX$ and $AP$ are isogonal. Using the same calculations as earlier, we can also get that $CX$ and $CP$ are isogonal, so $X$ exists. Similarly for $Y$, so we are done.

Toss the problem onto the complex plane, where the unit circle is circumcircle of $ABCD$, denoted by $\omega$. Let line $AX$ hit $\omega$ again at $M$ and $CX$ hit $\omega$ again at $N$. Then the angle conditions give us $m = \tfrac{bd}{c}$ and $n = \tfrac{bd}{ac}$. The chord intersection formula then gives $$\begin{aligned} x &= \frac{\tfrac{bdc}{a}(a + \tfrac{bd}{c}) - \tfrac{abd}{c}(c + \tfrac{bd}{a})}{\tfrac{bdc}{a} - \tfrac{abd}{c}} \\ &= \frac{c^2(a + \tfrac{bd}{c}) - a^2(c + \tfrac{bd}{a})}{c^2 - a^2} \\ &= \frac{ac + bd}{a + c}. \end{aligned}$$The condition that $X$ lies on $\overline{BD}$ then translates into $x + bd\overline{x} = b + d$, or $$\begin{aligned} \frac{ac + bd}{a + c} + \frac{bd(\tfrac{1}{ac} + \tfrac{1}{bd})}{\frac{1}{a} + \frac{1}{c}} &= b + d, \\ \frac{2ac + 2bd}{a + c} &= b + d ,\\ 2(ac + bd) &= (a + c)(b + d). \end{aligned}$$Since this stays the same under a rotation of the variables, it is the same condition that we find for $Y$, so we are done.

mymathboy wrote:

bro seriously wrote the sol b4 the thread was posted in his pm to himself, copy pasted, made an image, and sniped the post all within one minute. real skill

Benq wrote: I spent 4 hours on this and got nothing... Have you tried starting by solving for test cases 2-5

Wait this is an insanely stupid problem, especially for J5. I claim both conditions are equivalent to $AB \cdot CD = AD \cdot BC$. Proving one direction is enough. We notice the famous spiral similarity $\triangle AXD \sim\triangle ABC$ (which is featured in the proof of Ptolemy's theorem!), which implies that $$\frac{DX}{BC} = \frac{AD}{AC} \iff DX = \frac{AD \cdot BC}{AC}.$$Similarly, $\triangle CXD \sim \triangle CBA$, so $$\frac{AB \cdot CD}{AC} = DX = \frac{AD \cdot BC}{AC},$$hence the result.

i love complex numbers. Let the circumcircle of quadrilateral $ABCD$ be the unit circle on the complex plane. Let $A = a$, $B = b$, $C = c$, $D = d$. Note that since $X$ lies on $BD$, we have $x + bd\overline{x} = b+d$, meaning $\overline{x} = \frac{b+d-x}{bd}$. Then, from the first condition, $$\frac{\frac{a-d}{a-x}}{\frac{a-c}{a-b}} = \overline{\frac{\frac{a-d}{a-x}}{\frac{a-c}{a-b}}}$$$$\frac{\frac{a-d}{a-x}}{\frac{a-c}{a-b}} = \frac{\frac{\frac{1}{a} - \frac{1}{d}}{\frac{1}{a} - \frac{b+d-x}{bd}}}{\frac{\frac{1}{a} - \frac{1}{c}}{\frac{1}{a} - \frac{1}{b}}} = \frac{\frac{bd-ab}{bd-ab-ad+ax}}{\frac{bc-ab}{bc-ac}}$$$$\frac{(a-d)(a-b)}{(a-x)(a-c)} = \frac{b(d-a)(c)(b-a)}{b(c-a)(bd-ab-ac+ax)}$$$$-\frac{1}{a-x} = \frac{c}{bd-ab-ad+ax}$$$$bd-ab-ad+ax=cx-ac$$$$bd+ax+ac = ab+ad+cx.$$Now, we do the second condition. $$\frac{\frac{c-d}{c-x}}{\frac{c-a}{c-b}} = \overline{\frac{\frac{c-d}{c-x}}{\frac{c-a}{c-b}}}$$$$\frac{\frac{c-d}{c-x}}{\frac{c-a}{c-b}} = \frac{\frac{\frac{1}{c} - \frac{1}{d}}{\frac{1}{c} - \frac{b+d-x}{bd}}}{\frac{\frac{1}{c} - \frac{1}{a}}{\frac{1}{c} -\frac{1}{b}}} = \frac{\frac{bd-bc}{bd-cb-cd+cx}}{\frac{ab-bc}{ab-ac}}$$$$\frac{(c-d)(c-b)}{(c-x)(c-a)} = \frac{b(d-c)(a)(b-c)}{(bd-cb-cd+cx)(b)(a-c)}$$$$-\frac{1}{c-x} = \frac{a}{bd-cb-cd+cx}$$$$bd-cb-cd+cx = ax-ac$$$$bd+cx+ac = cb+cd+ax.$$Now, we place these two equations next to each other: $$bd+ax+ac=ab+ad+cx$$$$bd+cx+ac=cb+cd+ax$$and sum. $$2bd+2ac=ab+cb+ad+cd=(a+c)(b+d).$$This is same under the rotation of variables, and since all our steps are reversible, one true implies the other.

Is trig the intended solution here? feels very clean and straightforward We can rewrite the condition on the existence of $X$ as follows: if $X_1$ is the point on $\overline{BD}$ such that $\angle X_1 CD = \angle BCA$ and $X_2$ is the point on $\overline{BD}$ such that $\angle X_2 AD = \angle BAC$, then $DX_1 = DX_2$. Using law of sines, \[ DX_1 = DX_2 \Longleftrightarrow CD \sin \angle BCA = AD \sin \angle BAC \Longleftrightarrow AD = CD \times \frac{\sin \angle BCA}{\sin \angle BAC} \Longleftrightarrow AD = \frac{CD \times AB}{BC}.\] A similar process once again shows that the second condition is also equivalent to $AD = \frac{CD \times AB}{BC}$, so both conditions are equivalent.

@above agien proj geo is intended more clean and straightforward We will prove the if direction, from which the converse follows immediately from being the same config; namely, we'll show that Y is the midpoint of AC. Indeed, notice that $$CBY=ABD=YCD,YCB=ADB=YDC\implies\frac{BY}{CY}=\frac{CY}{DY}\iff CY^2=BY*DY,$$and analogously we can show AY^2=BY*DY, hence AY=CY. From here BX->BD is a symmedian of ABC, hence ABCD is harmonic; it follows that X is a midpoint since AX is median, and indeed, both ways hold to showing that ABCD is harmonic. $\blacksquare$

Since $\angle XDA = \angle ACB$ and $\angle CAB = \angle XAD$ we have $\triangle CAB \sim \triangle CDX$, and similarly $\triangle AXD \sim \triangle ABC$. From similar triangle ratios we get that $DX \cdot CA = AB \cdot CD$, and $XD \cdot AC = AD \cdot BC$, so $AB \cdot CD = AD \cdot BC$ so $ABCD$ is quasi-harmonic. Then it's well known that the lines isogonal to $BD$ wrt $\angle ADC$ and $\angle ABC$ concur at $Y$ since $BD$ is a symmedian wrt $\triangle ADC$ and $\triangle ABC$, so the existence of $Y$ is guaranteed. Analogously, the existence of $X$ also occurs given the existence of $Y$.

Use that one known length lemma about isogonals (and prove it by trig bash) to get $X$ exists iff $ABCD$ is harmonic, and same for $Y$. $\square$

Overkill sol for fun Note that $X$ is the $B$-Dumpty point of $\triangle ABC$ and the $D$-Dumpty point of $\triangle CDA$, so $BD$ is the symmedian implying \[ \dfrac{AB^2}{BC^2} = \dfrac{AP}{CP} = \dfrac{DA^2}{CD^2} \implies AB \cdot CD = BC \cdot DA \]where $P$ is the intersection of $AC$ and $BD$. But this relation is symmetric, which finishes.