ABCD IS tangential, if KLMN is cyclic then ABCD is cyclic

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ABCD IS tangential, if KLMN is cyclic then ABCD is cyclic

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Source: Iranian Geometry Olympiad 2018 IGO Advanced p4

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#1 h Sep 19, 2018, 7:52 PM • 1 Y

Y by Adventure10

Quadrilateral $ABCD$ is circumscribed around a circle. Diagonals $AC,BD$ are not perpendicular to each other. The angle bisectors of angles between these diagonals, intersect the segments $AB,BC,CD$ and $DA$ at points $K,L,M$ and $N$ . Given that $KLMN$ is cyclic, prove that so is $ABCD$ .

Proposed by Nikolai Beluhov (Bulgaria)

This post has been edited 2 times. Last edited by parmenides51, Sep 20, 2018, 9:25 AM
Reason: Proposed

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jbaca

#3 h Sep 19, 2018, 9:40 PM • 1 Y

Y by Adventure10

A sketch: In my opinion, this is a pretty hard but natural problem. Just use the angle bisectors to obtain two harmonic bundles, which easily gets to conclude that $KL$ , $MN$ and $AC$ concur at a point (possibly at infinity). Define $K'L'M'N'$ to be the quadrilateral formed by the tangency points of the incircle and sides $AB, BC, CD, DA$ respectively. A suitable application of Brianchon's and Pascal's theorems, and some harmonic bundles lead us to infer that $KL, K'L', AC, M'N', MN$ concur at a point $Q$ ; $LM, L'M', BD, KN, K'N'$ concur at $R$ and $KM, LN, K'M', L'N', AC, BD$ concur at $P$ . Using harmonic bundles again, we can show that $BD, AC$ are the $Q, R$ -polars with respect to the incircle and $(KLMN)$ , and from here it's not hard to argue that these circles must be concentric and finally, congruent, so $K=K'$ and the other equalities hold simultaneously. The final step is just to prove that $ABCD$ is cyclic, which is just a simple matter of angle chasing.

The no-perpendicularity constraint of $AC$ and $BD$ prevents us from having the extreme case of $KL\parallel MN, \ LM\parallel LN$ . The official solution includes hostile reasoning of how such a condition avoids the aforesaid parallelisms, which turns out to be trivial using harmonic bundles.

This post has been edited 2 times. Last edited by jbaca, Aug 4, 2023, 4:27 PM

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#4 h Sep 20, 2018, 6:25 AM • 2 Y

Y by guptaamitu1, Adventure10

Jafet98 wrote:

I've just realized that the no-perpendicularity constraint of $AC$ and $BD$ prevents us of having the extreme case of $KL\parallel MN, \ LM\parallel LN$ . Well, I didn't pay attention to this issue since, when using harmonic bundles (and the real projective plane), you just add a point at infinity to each class of parallel lines and the line at infinity as well. However, the official solution includes a hostile reasoning of how such a condition avoids the aforesaid parallelisms, which turns out to be trivial using harmonic bundles.

Sure, but the conclusion of the circumcircles of $KLMN$ and $K'L'M'N'$ being concentric doesn't hold when the concurrency points are ideal. In fact, the condition given in the problem trivially holds whenever $ABCD$ is a (not necessarily cyclic) kite. I do agree that the argument given in the official solution is unnecessarily complex though.

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#5 h Sep 20, 2018, 5:04 PM • 2 Y

Y by Adventure10, Mango247

juckter wrote:

Jafet98 wrote:

I've just realized that the no-perpendicularity constraint of $AC$ and $BD$ prevents us of having the extreme case of $KL\parallel MN, \ LM\parallel LN$ . Well, I didn't pay attention to this issue since, when using harmonic bundles (and the real projective plane), you just add a point at infinity to each class of parallel lines and the line at infinity as well. However, the official solution includes a hostile reasoning of how such a condition avoids the aforesaid parallelisms, which turns out to be trivial using harmonic bundles.

Sure, but the conclusion of the circumcircles of $KLMN$ and $K'L'M'N'$ being concentric doesn't hold when the concurrency points are ideal. In fact, the condition given in the problem trivially holds whenever $ABCD$ is a (not necessarily cyclic) kite. I do agree that the argument given in the official solution is unnecessarily complex though.

I agree with you. I'm now left to wait the final results.

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#6 h May 30, 2019, 6:14 AM • 6 Y

Y by Adventure10, UI_MathZ_25, Mango247, Mango247, Mango247, Stuffybear

Let $E=\overline{AB}\cap\overline{CD}$ , $X=\overline{AB}\cap\overline{CD}$ , $Y=\overline{AD}\cap\overline{BC}$ . Also let the incircle of $ABCD$ touch $\overline{AB}$ , $\overline{BC}$ , $\overline{CD}$ , $\overline{DA}$ at $K'$ , $L'$ , $M'$ , $N'$ . I claim $K=K'$ , etc.

Claim: $\overline{KL}\cap\overline{MN}$ is the harmonic conjugate of $E$ wrt.\ $\overline{AC}$ . (In particular, it lies on $\overline{AC}$ .)

Proof. Let $K^*=\overline{AB}\cap\overline{LN}$ and $L^*=\overline{BC}\cap\overline{KM}$ . From the angle bisectors we have $-1=(AB;KK')=(CB;LL')$ , so $\overline{AC}$ , $\overline{KL}$ , $\overline{K^*L^*}$ concur at a point $F$ . But $-1=(AB;KK^*)\stackrel{L^*}=(AC;EF).$ The claim readily follows by symmetry. $\blacksquare$

$[asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=red; pen sec=orange; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair I, K, L, M, NN, A, B, C, D, EE, F, G, X, Y, Kp, Lp; I=(0, 0); K=dir(120); L=dir(40); M=dir(270); NN=2*foot(I, L, foot(L, K, M))-L; A=2*(NN+K)/length(NN+K)^2; B=2*(K+L)/length(K+L)^2; C=2*(L+M)/length(L+M)^2; D=2*(M+NN)/length(M+NN)^2; EE=extension(K, M, L, NN); F=extension(K, L, M, NN); G=extension(K, NN, L, M); X=extension(A, B, C, D); Y=extension(A, D, B, C); Kp=extension(A, B, L, NN); Lp=extension(B, C, K, M); filldraw(circumcircle(A, B, C), fil, pri); fill(A -- B -- C -- D -- cycle, fil); filldraw(circle(I, 1), sfil, sec); fill(K -- L -- M -- NN -- cycle, sfil); draw(F -- L -- M -- F, sec); draw(L -- M -- NN -- K, sec); draw(X -- B -- C -- X, pri); draw(Lp -- C -- D -- Y, pri); draw(Y -- X, tri); draw(F -- C, tri); draw(B -- D, tri); draw(Lp -- M, sec); draw(Kp -- L, sec); draw(Kp -- Lp,sec); dot("\(A\)", A, dir(100)); dot("\(B\)", B, dir(70)); dot("\(C\)", C, SE); dot("\(D\)", D, SW); dot("\(K\)", K, dir(75)); dot("\(L\)", L, NE); dot("\(M\)", M, S); dot("\(N\)", NN, dir(240)); dot("\(X\)", X, SW); dot("\(Y\)", Y, NE); dot("\(F\)", F, NW); dot("\(E\)", EE, dir(295)); dot("\(K^*\)", Kp, dir(300)); dot("\(L^*\)", Lp, N); [/asy]$

Claim: $E=\overline{K'M'}\cap\overline{L'N'}$ .

Proof. Brianchon theorem on $AK'BCM'D$ , $ABL'CDN'$ . $\blacksquare$

Claim: $\overline{K'L'}\cap\overline{M'N'}$ is the harmonic conjugate of $E$ wrt.\ $\overline{AC}$ . (In particular, it lies on $\overline{AC}$ .)

Proof. By Pascal theorem on $K'K'L'N'N'M'$ , the point $F=\overline{K'L'}\cap\overline{M'N'}$ lies on $\overline{AEC}$ . Also let $G=\overline{K'N'}\cap\overline{L'M'}$ . Note that $-1=G(K'L';F'E')=(AC;EF),$ where the second equality is by taking the dual wrt.\ $(K'L'M'N')$ . $\blacksquare$

Remark: By Pascal theorem on $K'K'L'M'M'N'$ and $K'L'L'M'N'N'$ , points $X$ , $Y$ lie on $\overline{FG}$ .

Claim: $K=K'$ , $L=L'$ , $M=M'$ , $N=N'$ .

Proof. We have proven via Claims 2 and 3 that $(KLMN)$ and $(K'L'M'N')$ coincide as the polar circle of $\triangle EXY$ . But $(K'L'M'N')$ is tangent to each side of $ABCD$ , so $K=K'$ is unique, etc. $\blacksquare$

Finally, $\overline{KM}\cap\overline{LN}$ readily implies $ABCD$ is bicentric; to spell it out, $\angle BAD=180^\circ-\widehat{NK}=\widehat{LM}=180^\circ-\angle DCB,$ as needed.

This post has been edited 2 times. Last edited by TheUltimate123, May 27, 2020, 9:37 PM

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