IMO 2012 P5

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1547 posts

#1 h Jul 11, 2012, 7:03 PM • 30 Y

Y by Pirshtuk, siavosh, buratinogigle, Potla, KOSNITA, Amir Hossein, samuelbf, Mathematicalx, Scorpion.k48, tenplusten, Davi-8191, anantmudgal09, Wizard_32, nguyendangkhoa17112003, XbenX, Gumnaami_1945, Richangles, amar_04, Adventure10, Mango247, home245, GeoKing, Rounak_iitr, Tastymooncake2, crazyeyemoody907, ItsBesi, Sedro, and 3 other users

Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$ , and let $D$ be the foot of the altitude from $C$ . Let $X$ be a point in the interior of the segment $CD$ . Let $K$ be the point on the segment $AX$ such that $BK=BC$ . Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$ . Let $M$ be the point of intersection of $AL$ and $BK$ .

Show that $MK=ML$ .

Proposed by Josef Tkadlec, Czech Republic

This post has been edited 3 times. Last edited by Eternica, Jun 19, 2024, 10:03 AM

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Concyclicboy

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Concyclicboy

#2 h Jul 11, 2012, 7:46 PM • 17 Y

Y by siavosh, buratinogigle, Swistak, armann, ZacPower123, myh2910, Jalil_Huseynov, Adventure10, Mango247, Tastymooncake2, Sedro, and 6 other users

I think:
1° AL^2=AD*AB then <ALD=<LBA and a tangent to circuncircles is obviously
2° Let F the intersection of CD and the perpendicular to AL through L, then FLDA is cyclic and <DFA=<DLA=<LBA
3 If we call T the intersection of BX with AF then BFTD is cyclic and X is orthocenter with AX perpendicular to BF

By similar way the perpendicular for BK through K passes through F, and if U is the intersection of AX with BF we have: FL^2=FT*FA=FU*FB=FK^2, impliying that LM=KM

This post has been edited 1 time. Last edited by Concyclicboy, Jul 12, 2012, 1:19 AM

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Aryan5000

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Aryan5000

#3 h Jul 11, 2012, 7:55 PM • 13 Y

Y by siavosh, ropro01, buratinogigle, Swistak, Algie, myh2910, Adventure10, Tastymooncake2, and 5 other users

MK and ML are both tangents to a circle
let BX meets the circle(A,AC)at J And AX meets circle(B,BC)at I(circle (c,r)means circle with center c and radius r)
easily we can find that JKLI is cyclic Let name this that have these four points on itself circle C2
easily we can find that BK is tangent to C2 (because we have BC*BC=Bl*BJ and we have BC =BK
so BK*BK=BL*BJ)
with a similar work we can see that AL is tangent to the C2 too.
So MK and ML are both tangents to the C2 SO MK=ML

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Potla

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Potla

#4 h Jul 11, 2012, 8:07 PM • 14 Y

Y by buratinogigle, diablo901, mathbuzz, Sx763_, biomathematics, Gumnaami_1945, Adventure10, Tastymooncake2, MS_asdfgzxcvb, and 5 other users

Obviously the perpendiculars through $K, L$ and to $BK, AL$ intersect at a point $E$ which lies on $CD,$ which is also the intersection of $\odot BDK$ and $\odot ADL.$ So, by some angle chasing, it follows that $\angle EKD=\angle AXE;$ meaning that $EK$ is tangent to $\odot KDX.$ So, by power of $E$ , it follows that $EK^2=ED\cdot EX=EL^2,$ because $EL$ is also tangent to $\odot LDX.$ Therefore $EK=EL,$ and $MKE\cong MLE,$ leading to $MK=ML.\Box$

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buratinogigle

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buratinogigle

#5 h Jul 11, 2012, 8:12 PM • 10 Y

Y by silly_mouse, phuongtheong, p1a, PARISsaintGERMAIN, Adventure10, Mango247, Tastymooncake2, and 3 other users

Let me propose this general problem

Let two circles $(O_1)$ and $(O_2)$ intersect at $A,B$ . $C,D$ are on line $O_1O_2$ such that $AC\perp O_1A$ and $AD\perp O_2A$ . $P$ is a point on $AB$ . $CP$ cut $(O_1)$ at $L$ such that $C,L$ don't have same side with $AB$ . $DP$ cut $(O_2)$ at $K$ such that $D,K$ don't have same side with $AB$ . $LO_1$ cuts $KO_2$ at $M$ . Prove that $MK=ML$ .

Attachments:

Figure856.pdf (6kb)

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Goutham

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Goutham

#6 h Jul 11, 2012, 8:56 PM • 12 Y

Y by buratinogigle, p1a, Wizard_32, Adventure10, Mango247, Tastymooncake2, and 6 other users

Let circumcircle of $ADL$ intersect $DC$ at $U$ . Then $\angle AUD=\angle ALD$ . Also, $AL^2=AC^2=AD\cdot AB$ and so, $LA$ is antiparallel to $BD$ which means, $\angle AUD=\angle LBD=\angle XBD$ . Now, this implies $UAD\sim BXD$ . This implies $\frac{XD}{AD}=\frac{BD}{UD}\Longrightarrow \frac{UD}{BD}=\frac{AD}{XD}$ . So, $UDB\sim AXD$ . This means $\angle BUD=\angle DAX$ but by similar reasonings, $\angle DAX=\angle DKB$ . So, $BDKU$ is cyclic. Since in circles $ADLU$ and $BDKU$ , we have $AU, BU$ as diameters, we also have $\angle ALU=\angle BKU=\frac{\pi}{2}$ . But $U$ also lies on $CD$ . So, the perpendiculars from $K, L, D$ to $BM, AM, AB$ concur at a point, $U$ . By carnot's theorem, $BK^2-KM^2+ML^2-LA^2+AD^2-DB^2=0$ . Using $AL=AC, BK=BC, AD^2-DB^2=(AD^2+DC^2)-(DC^2+DB^2)=AC^2-CB^2$ , we have $MK=ML$ as required.

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vladimir92

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vladimir92

#7 h Jul 11, 2012, 9:48 PM • 7 Y

Y by diablo901, buratinogigle, p1a, Adventure10, Mango247, Tastymooncake2, and 1 other user

Let AX and BX meet the circumcircle of ABC at V and U respectively, AU and BV meet at W, clearly X is the orthocenter of WAB. See that AL²=AD.AB=AX.AV, then A,D,L,V,W are concyclic, similarly B,D,K,U,W are concyclic too. See that WLV~WBL and WKU~WAK, it follows that WL=WK, but WLKM cyclic, hence ML=MK, As desired.

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Goutham

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Goutham

#8 h Jul 11, 2012, 9:52 PM • 9 Y

Y by buratinogigle, p1a, PARISsaintGERMAIN, IFA, Adventure10, Mango247, Tastymooncake2, and 2 other users

buratinogigle wrote:

Let me propose this general problem

Let two circles $(O_1)$ and $(O_2)$ intersect at $A,B$ . $C,D$ are on line $O_1O_2$ such that $AC\perp O_1A$ and $AD\perp O_2A$ . $P$ is a point on $AB$ . $CP$ cut $(O_1)$ at $L$ such that $C,L$ don't have same side with $AB$ . $DP$ cut $(O_2)$ at $K$ such that $D,K$ don't have same side with $AB$ . $LO_1$ cuts $KO_2$ at $M$ . Prove that $MK=ML$ .

Let $AB$ intersect $CD$ at $S$ . Then, circumcircle of $O_1SL$ intersect $AB$ at $U$ . So, $\angle O_1US=\angle O_1LS$ . Also, $O_1L^2=O_1A^2=O_1S\cdot O_1C$ and so, $O_1L$ is tangent to circumcircle of $LSC\Longrightarrow \angle O_1LS=\angle LCS=\angle PCS$ . So, $PCS\sim O_1US\Longrightarrow \frac{PS}{O_1S}=\frac{CS}{US}$ . Now, $O_1AS\sim ACS\Longrightarrow AS^2=O_1S\cdot CS$ . Similarly, $AS^2=O_2S\cdot DS$ . This implies $\frac{O_1S}{O_2S}=\frac{DS}{CS}$ . Multiplying this with $\frac{PS}{O_1S}=\frac{CS}{US}$ , we have $\frac{PS}{O_2S}=\frac{DS}{US}\Longrightarrow PDS\sim O_2US$ . So, $\angle O_2US=\angle PDS$ . We also have $O_2K^2=O_2A^2=O_2S\cdot O_2D$ and hence, $O_2K$ is tangent to cirumcircle of $KSD\Longrightarrow \angle PDS=\angle KDS=\angle O_2KS$ . This means $O_2KUS$ is cyclic. In circles $O_1LUS$ and $O_2KUS$ , we have $O_1U$ and $O_2U$ as diameters and so, $\angle O_1LU=\angle O_2KU=\frac{\pi}{2}$ . $K$ lies on $AB$ . The perpendiculars to sides $MO_1, MO_2, O_1O_2$ passing through $L, K, S$ concur at $U$ . Hence, by carnot's theorem, $ML^2-LO_1^2+O_1S^2-SO_2^2+O_2K^2-KM^2=0$ . We have $O_1S^2-O_2S^2=(O_1S^2+SA^2)-(SA^2+O_2S^2)=O_1A^2-O_2A^2=O_1L^2-O_2K^2$ . Using this, we have $MK=ML$ as desired.

When $\angle O_1AO_2=\frac{\pi}{2}$ , then $D, C$ coincide with $O_1, O_2$ respectively. In that case, it can be seen that we get the IMO problem.

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pavel kozlov

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pavel kozlov

#9 h Jul 12, 2012, 9:39 AM • 2 Y

Y by Adventure10, Mango247

Another addition to this problem:
Prove that the bisector of $\angle{ACB}$ and the lines $XM$ and $AB$ meet each other at one point.

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pavel kozlov

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pavel kozlov

#10 h Jul 12, 2012, 9:54 AM • 3 Y

Y by Goutham, Adventure10, Mango247

I very like the solution of Goutham with Carnot's theorem. Very effective finalization really!
By the way: accurate calculation in Cartesian coordinates together with the sines theorem applied to triangles $AKM, BLM, ALX, BKX$ (that trick excludes calculation of the point $M$ ) gives desire result in half an hour.

This post has been edited 1 time. Last edited by pavel kozlov, Jul 14, 2012, 10:40 PM

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ropro01

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ropro01

#11 h Jul 12, 2012, 12:43 PM • 11 Y

Y by dizzy, Hantaehee, TheMathsBoy, Jalil_Huseynov, Adventure10, Mango247, Tastymooncake2, Rounak_iitr, and 3 other users

Very nice problem, my favorite one this year

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Orin

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Orin

#12 h Jul 12, 2012, 12:53 PM • 6 Y

Y by numbertheorist17, Adventure10, Mango247, Tastymooncake2, and 2 other users

First we construct three circles:the circumcircles of $\triangle ABC(\Gamma),$ ,The circle( $\alpha$ ) with centre $A$ and radius $AC$ and the circle( $\beta$ ) with centre $B$ and radius $BC$ .
Note that the centre of $\Gamma$ lies on the midpoint of $AB$ ,so the three circles are co-axial with radical axis $CD$ .
Let $AX \cap \Gamma =Y \neq A,BX \cap \Gamma =Z \neq B,AY \cap BZ =P$ .
so $\angle AYB =\angle AZB=90^{\circ}$ ,so $X$ must be the orthocenter of $\triangle ABP$ implying $P$ lies on the radical axis.
let us denote $PWR(P)$ by the power of $P$ wrt $\alpha,\beta,\Gamma$ .[which are equal]
From similar triangles $ABC,ACD$ we get $AC^2=AD*AB=AL^2 \Rightarrow \angle ALD = \angle ABL$
similarly $\angle BKD = \angle BAK$
now $\angle APD=\angle ABZ=\angle ALZ$ implying $ADLP$ is cyclic.
so $AL \bot PL$ .....(1) meaning $PL^2=PWR(P)$ .
similarly we get $BK \bot PK$ .....(2) meaning $PK^2=PWR(P)$
hence $PK=PL$ .....(3)
so $\triangle MKP \cong MLP$ [using (1),(2),(3)]
proving $MK=ML$

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Jeroen

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Jeroen

#13 h Jul 12, 2012, 3:29 PM • 6 Y

Y by buratinogigle, Swistak, Adventure10, Mango247, Tastymooncake2, and 1 other user

This was the NLD6's solution:

$E$ is the reflection of $C$ in $AB$ . It lies on the circle $\Gamma$ with centre $B$ en radius $BC$ , as does $K$ . The line $AK$ intersects $\Gamma$ a second time in $K'$ and the tangent in $K$ to $\Gamma$ intersects $CD$ in $S$ . The tangents in $C$ and $E$ to $\Gamma$ intersect in $A$ en $A$ is on $KK'$ , so $CEKK'$ is a harmonic quadrilateral. This yields $-1=(CEKK')=K(CEKK')=K(CESX)=(CESX)$ .

If we define $T$ to be the point of intersection of $CD$ and the tangent in $L$ to the circle with centre $A$ and radius $AC$ , we similarly find $(CETX)=-1$ . From this follows $S=T$ and the power of a point now yields $SK^2=SC\cdot SE=SL^2$ . So $SK=SL$ and now we have congruent triangles $SKM$ and $SLM$ , which yields $KM=LM$ .

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proglote

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proglote

#14 h Jul 12, 2012, 3:34 PM • 11 Y

Y by buratinogigle, TheStrangeCharm, like123, Roct-7, Adventure10, Mango247, Tastymooncake2, and 4 other users

Denote by $\mathcal{C}_A$ the circle centered at $A$ passing through $C$ , and define $\mathcal{C}_B$ similarly. Suppose $AK$ intersects $\mathcal{C}_B$ again at $T_A$ , and define $T_B$ similarly. Observe that $X$ is in the radical axis $\ell$ of $\mathcal{C}_A, \mathcal{C}_B$ , so $XL \cdot XT_B = XK \cdot XT_A \implies KLT_AT_B$ cyclic with circumcircle $\omega$ . Note that $\ell$ is the polar of $A$ w.r.t. $\mathcal{C}_B$ , so $(A, X; K, T_A)$ is harmonic $\implies X$ is in the polar $a$ of $A$ w.r.t. $\omega$ . But $AL = AT_B \implies a \parallel LT_B \implies a \equiv LT_B.$ Similarly $b \equiv KT_A.$

So $AL, BK$ are tangent to $\omega.$ So $MK, ML$ are tangent to $\omega.$ So done.

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Neerjhor

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Neerjhor

#15 h Jul 13, 2012, 8:08 AM • 6 Y

Y by p1a, Jalil_Huseynov, Adventure10, Tastymooncake2, and 2 other users

I found this problem easy. I'm very confused as everyone saying it's too hard. Please tell me if there's a bug. :S

Let $AX$ and $BX$ intersect the circumcircle $O_1$ of $ABC$ at $P$ and $Q$ . Let $AQ$ and $BP$ intersect at $S$ . By Radical Axis theorem (I mean Radical Centre), $S$ lies on $CD$ . Now, triangle $ABC$ and $ACD$ are similar. So, $AC^2=AB.AD=AL^2$ . Again, as quadrilateral $SQDB$ is cyclic, so $AD.AB=AQ.AS=AL^2$ . This means that $AL$ is a tangent to the circumcircle of triangle $SQL$ at $L$ . And so, $AL$ is perpendicular to $SL$ as $SQL=90^0$ . Similarly, $BK$ is perpendicular to $SK$ . And $SK=SL$ because they lie on the radical axis of circle $(A,AC)$ and circle $(B,BC)$ and are tangents to them. So, if we draw a circle with centre $S$ and radius $SK$ then it would go through $K,L$ and as $<MKS=90^0$ and $<MLS=90^0$ , it implies $MK=ML$ .

I hope I got it right. :/

This post has been edited 1 time. Last edited by Neerjhor, Jul 13, 2012, 1:33 PM

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peace09

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peace09

#114 h Jun 25, 2023, 8:01 PM

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'nother fact that holds true:

Claim. $\tfrac{C_DL}{CL}=\tfrac{C_DK}{CK}$ , where $C_D$ is the reflection of $C$ across $AB$ .

Proof. Let $\omega_A,\omega_B$ be the circles centered at $A$ passing through $C,L$ and centered at $B$ passing through $C,K$ , respectively. Additionally, define $K_2,L_2$ to be the second intersections of $BK,AL$ with $\omega_A,\omega_B$ respectively, and for completeness' sake add $C_A,C_B$ , the antipodes of $C$ with respect to $\omega_A,\omega_B$ .

The trick is to invert at $C$ to produce the following (awfully low resolution, sorry) diagram:

https://cdn.artofproblemsolving.com/attachments/e/0/b8aaaae6a8cae5288aee2c656e87225910cd25.png

Here, $\omega_A'\perp\omega_B'$ implies that $C_D'$ is the midpoint of both $K'K_2'$ and $L'L_2'$ . But $C_D'$ lies on the radical axis of $(A'CX')$ and $(B'CX')$ , meaning
$C_D'K'{}^2=C_D'K'\cdot C_D'K_2'=C_D'L'\cdot C_D'L_2'=C_D'L'{}^2.$ So $C_D'K'=C_D'L'$ , which becomes $\tfrac{C_DK\cdot r^2}{CC_D\cdot CK}=\tfrac{C_DL\cdot r^2}{CC_D\cdot CL}$ where $r$ is the radius of inversion. Rearranging gives the desired. $\Box$

This post has been edited 1 time. Last edited by peace09, Jun 25, 2023, 8:01 PM

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minusonetwelth

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minusonetwelth

#115 h Sep 18, 2023, 8:16 PM

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The main idea is to see $MK$ and $ML$ as tangents to some circle. Call $\gamma_A$ the circle through $A$ with radius $AC$ and define $\gamma_B$ similarly. Let $\gamma=(ABC)$ . Define $L'=BX\cap \gamma_A$ and $K'=AX\cap \gamma_B$ . Let $C'$ be the reflection of $C$ over $AB$ , which clearly lies on $\gamma$ , $\gamma_A$ and $\gamma_B$ . Then
$XK'\cdot XK=XC\cdot XC'=XL\cdot XL',$ so $L'KLK'$ is cyclic. Then, $BK^2=BC^2=BL\cdot BL'$ , so $BK$ and thus $MK$ is tangent to $(L'KLK')$ . Similarly, $ML$ is tangent to this circle, implying the result by equal tangents.

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cursed_tangent1434

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cursed_tangent1434

#116 h Dec 7, 2023, 12:46 AM

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Let $S$ be the reflection of $C$ over $AB$ . Since $\measuredangle ASB = 90^\circ = \measuredangle ACB$ , it is easy to see that $S$ lies on $(ABC)$ . Now, let $\omega_K$ be the circle centered at $B$ with radius $BK$ and $\omega_L$ the circle centered at $A$ with radius $AL$ . Since,
$AC=AL=AS \text{ and } BC=BK=BS$ we also have that $C$ and $S$ lie on both $\omega_B$ and $\omega_C$ . Let $K'= \overline{AK} \cap \omega_K \neq K$ and $L' = \overline{BL} \cap \omega_L \neq L$ . Since,
$XL\cdot XL' = XC \cdot XS = XK \cdot XK'$ we also have that $KLK'L'$ is cyclic (let its circumcircle be $\Gamma$ ). Now, note that
$BL \cdot BL' = \text{Pow}_{\omega_L}(B)= BA^2-AC^2 = BC^2$ Thus, $BC$ is a tangent to $\omega_L$ . Similarly, $AC$ is a tangent to $\omega_K$ . Now, note that,
$BK^2=BC^2=BL \cdot BL'$ Thus, $BK$ is a tangent to $\Gamma$ . Similarly, $AL$ is also a tangent to $\Gamma$ . Now, this means, that $M$ is the intersection of the tangents to $\Gamma$ at $K$ and $L$ and thus, we have
$MK=ML$ which was the required conclusion.

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stevenmeow

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stevenmeow

#117 h Dec 23, 2023, 12:34 AM

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Here is a cheesy computational solution which uses is nearly limited to addition and multiplication. This necromantic post is partly for my own benefit. The following is probably hard to find compared to other computational solutions here. We use the notation aa for the square of a, following the style of Riemann (see wiki: On_the_Number_of_Primes_Less_Than_a_Given_Magnitude )

In fact, we are using X as the main angle. Define $a = AX, b = BX, k = KX, l = LX$ . Let $c = -\cos(\angle AXB)$ . Note that $[X, K, A]$ are collinear in that order as are $[X, L, B]$ .

[1] $AL^2 = aa+ll+2alc.$
[2] $BK^2 = bb+kk+2bkc.$
[3] Given that $AL^2 + BK^2 = AC^2 + BC^2 = AB^2 = aa+bb+2abc,$
[4] $ll+kk=2c(ab-al-bk).$
From a standard lemma about perpendicular lines and lengths applied to $AB$ and $CX$ ,
[5] $aa + BK^2 = AX^2 + BC^2 = BX^2 + AC^2 = bb + AL^2.$
Substituting,
[6] $kk-ll=2c(al-bk).$

Now working backwards, it suffices to show the following, using the law of sines on triangles MAK and MBL (excuse the fractions)
[7 STS] $AK / \sin( \angle AMK ) * \sin ( \angle MAK ) = BL / \sin ( \angle BML ) * \sin ( \angle MBL ).$
We cancel the vertical angles, substitute (a, b, k, l), and use that a > k; b > l
[8 STS] $(a-k) \sin ( \angle LAX ) = (b-l) \sin( \angle BKX ).$
Using the the law of sines on triangles LAX and BKX,
[9 STS] $(a-k) l \sin ( \angle X) / AL = (b-l) k \sin( \angle X) / BK.$
Since both sides are positive, we can square both sides. Also we cancel out sin(X), clear denominators (I was getting worried about the fractions there) and substitute [1] and [2].
[10 STS] $(a-k)^2 ll (bb + kk + 2bkc) = (b-l)^2 kk (aa+ll-2alc).$

[4 + 6] $kk = abc - 2bkc.$
[4 - 6] $ll = abc - 2alc.$

We use [4 + 6] and [4 - 6] to replace kk and ll in [10] (but not in $(a-k)^2, (b-l)^2$ ).
[11 STS] $(a-k)^2 (abc - 2alc) (bb+abc) = (b-l)^2 (abc - 2bkc) (aa+abc).$
Cancelling factors of a, b, and c,
[12 STS] $(a-k)^2 (b-2l) (b+ac) = (b-l)^2 (a-2k) (a+bc).$

Here is the computational simplification that comes out of nowhere for a true geometry problem.

We observe $(b-2al)(b+ac) = bb-2bl+abc-2alc = bb-2bl=ll = (b-l)^2,$ using [4-6]. Similarly, $(a-2k)(a+bc) = (a-k)^2,$ so [12] is true and we are done.

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IAmTheHazard

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IAmTheHazard

#118 h Dec 27, 2023, 10:32 PM

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Let $\omega_A$ and $\omega_B$ be the circles centered at $A$ and $B$ (resp.) passing through $C$ , so $K$ and $L$ lie on $\omega_B$ and $\omega_A$ respectively. Let $K'$ and $L'$ be the other intersections of $\overline{AX}$ with $\omega_B$ and $\overline{BX}$ with $\omega_A$ .

By the converse of the radical axis theorem, $KLK'L'$ is cyclic. Furthermore, since $\angle BCA=90^\circ$ , $\omega_A$ and $\omega_B$ are orthogonal, so inverting at $A$ swaps $K$ and $K'$ (and obvious fixes $L$ and $L'$ ). Thus $\omega_A$ and $(KLK'L')$ are orthogonal, so by definition $\overline{AL}$ is tangent to $(KLK'L')$ . Likewise, $\overline{BK}$ is tangent to $(KLK'L')$ , hence $M$ is the intersection of the tangents from $K$ and $L$ to $(KLK'L')$ and $MK=ML$ . $\blacksquare$

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IAmTheHazard

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IAmTheHazard

#119 h Dec 27, 2023, 10:38 PM

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CyclicISLscelesTrapezoid wrote:

My solution is very similar to everyone else's, but I want to note something that I think is really cool. Let $\overline{AK}$ intersect the circle centered at $B$ through $C$ again at $P$ , and let $\overline{BL}$ intersect the circle centered at $A$ through $C$ again at $Q$ . Notice that lines $KQ$ and $LP$ intersect on $\overline{AB}$ , as do lines $KL$ and $PQ$ . There's a projective proof to this, but I challenge you to find the synthetic proof. Hint

Given the tangency facts above, this is just Brocard on $KLK'L'$ since $A$ is the pole of $\overline{KK'}$ and $B$ is the pole of $\overline{LL'}$ , so any proof of Brocard will work here and any proof of this should just prove Brocard's theorem

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trk08

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trk08

#120 h Jan 9, 2024, 2:28 AM

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Note that $K$ lies on the circle with center $B$ that goes through $C$ . Also, $L$ lies on the circle with center $A$ that goes through $C$ . Call these circles $\omega, \gamma$ , respectively. Also, let $T=(BDK)\cap (ADL)$

Claim:
$X$ is the radical center of $\omega, \gamma, (BDK), (ADL)$ .
Proof:
Note that $AC$ is tangent to $\omega$ . Therefore, using the given right angle condition:
$Pow_{\omega}(A)=AC^2=AD\cdot AB=Pow_{(BDK)}(A),$ so $AKX$ is the radical axis of $\omega$ and $(BDK)$ .

Similarly, $BLX$ is the radical axis of $(LDA)$ and $\gamma$ . Therefore, the radical center is:
$AKX\cap BLX=X$ $\square$

Claim:
$T$ lies on the radical axis of $\omega,\gamma$
Proof:
It suffices to show that $X$ lies on $TD$ . Note that $TD$ is the radical axis of $(ADL),(BDK)$ . Using our first claim, we are done $\square$

Claim:
The circle with center $T$ and goes through $L$ , goes through $K$ , and is tangent to $ML$ and $MK$
Proof:
Note:
$\angle TLA=\angle TDA=\angle TKB=90.$ Therefore:
$Pow_{\omega}(T)=TK^2=Pow_{\gamma}(T)=TL^2,$ so $TL=TK$ , implying that the circle exists. As $\angle TLM=\angle TKM=90$ , the tangency condition is implied $\square$

Therefore, $ML=MK$ , as desired $\blacksquare$

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shendrew7

788 posts

shendrew7

#121 h Jan 12, 2024, 4:26 AM

Y by

Let $K^*$ and $L^*$ be the inverses of $K$ and $L$ wrt $\omega_A$ and $\omega_B$ , the circles through $C$ centered at $A$ and $B$ , respectively.

$\omega_A$ and $\omega_B$ are orthogonal, so $K^*$ lies on $\omega_B$ and $L^*$ lies on $\omega_A$ .
$X$ lies on the radax of $\omega_A$ , and $\omega_B$ , so $KLK^*L^*$ is cyclic. Let its circumcircle be $\gamma$ .
$KLK^*L^*$ is also harmonic, with tangents from $K$ and $K^*$ meeting at $B$ and tangents from $L$ and $L^*$ meeting at $A$ , as
$AK \cdot AK^* = AC^2 = AL^2.$

Thus $MK$ and $ML$ are tangents to $\gamma$ , which finishes. $\blacksquare$

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ihatemath123

3427 posts

ihatemath123

#122 h Mar 8, 2024, 4:31 AM

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Let $\omega_A$ be the circle centered at $A$ passing through $C$ and $L$ ; define $\omega_B$ likewise. Clearly, $\omega_A$ and $\omega_B$ are orthagonal. Let $A_1$ be the second intersection of line $KX$ with $\omega_A$ ; define $B_1$ likewise. Let $L_1$ and $K_1$ be the second intersections of line $LK$ with $\omega_A$ and $\omega_B$ , respectively.

Inversion about $W_A$ sends $L_1$ and $L$ to itself, as well as sending $K$ to $A_1$ . Hence, $ALA_1L_1$ is cyclic, so
$\angle ALL_1 = \angle AL_1L = \angle AA_1L$ Similarly, $\angle BKK_1 = \angle BB_1K$ .

But by the radical axis theorem, $ABA_1B_1$ is cyclic, so $\angle AA_1L = \angle BB_1K$ , hence $\angle ALL_1 = \angle BKK_1$ . So, $\triangle MKL$ is isosceles as desired.

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CyclicISLscelesTrapezoid

371 posts

CyclicISLscelesTrapezoid

#124 h Jun 21, 2024, 3:21 PM • 1 Y

Y by GeoKing

Let $\omega_A$ and $\omega_B$ be the circles through $C$ centered at $A$ and $B$ , and let $C'$ be the reflection of $C$ over $\overline{AB}$ . Choose $Y$ on $\overline{CC'}$ such that $(C,C';X,Y)=-1$ . By projecting through $K$ , we have $(C,C';\overline{KX} \cap \omega_B,\overline{KY} \cap \omega_B)=-1$ , so $\overline{KY}$ is tangent to $\omega_B$ . Analogously, $\overline{LY}$ is tangent to $\omega_C$ .

By radical axis, there exists a circle tangent to $\omega_A$ at $L$ and $\omega_B$ at $K$ . The center of this circle lies on $\overline{AL}$ and $\overline{BK}$ , so it must be $M$ . Therefore, we have $MK=ML$ , as desired. $\square$

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SenorSloth

37 posts

SenorSloth

#125 h Jun 21, 2024, 6:37 PM • 1 Y

Y by OronSH

Diagram

Let $P$ be the foot of the perpendicular from $B$ to $AX$ , and let $Q$ be the foot of the perpendicular from $A$ to $BX$ . Let the circle centered at $B$ through $C$ be $\omega_1$ , and let the circle centered at $A$ through $C$ be $\omega_2$ . Define $Y$ to be the second intersection of $AX$ with $\omega_1$ , and let $Z$ be the second intersection of $BX$ with $\omega_2$ .

Note that $P$ and $Q$ lie on $(ABC)$ because of the right angles. Let $F=BP\cap AQ$ . We claim that $F$ lies on line $XD$ . By Power of a Point, $FP\cdot FB = FQ\cdot FA$ , which means $F$ has equal power to the circles $(BDXP)$ and $(ADXQ)$ . Thus, $F$ must lie on their radical axis, which is the line $XD$ .

Since $X$ lies on the radical axis of $\omega_1$ and $\omega_2$ , $LX\cdot XZ = KX \cdot XY$ , which implies that $KLYZ$ is cyclic. Note that since $F$ is the intersection of two perpendicular bisectors of sides of $KLYZ$ , it must be the center.

Note that $BP\cdot BF= BD\cdot BA=BC^2=BK^2$ , so $BK\perp KF$ . Similarly, $AQ\cdot AL=AD\cdot AB=AC^2=AL^2$ , so $AL\perp LF$ . Thus $\angle FLM = \angle FKM = 90^{\circ}$ . We also know $FL=FK$ since $F$ is the center of $(KLYZ)$ , and the side $FM$ is shared, so then $FLM\cong FKM$ and $MK=ML$ , as desired.

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Ywgh1

136 posts

Ywgh1

#126 h Aug 13, 2024, 8:39 AM

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IMO 2012 p5
Let $\omega_1$ and $\omega_2$ be circles Centered at $A$ and $B$ passing through $C$ .
Let $K’$ and $L’$ be the intersections of $AK$ and $BL$ with circles $\omega_1$ and $\omega_2$ .
By radical axis we have that $KLK’L’$ is cyclic.

Now since $\omega_1$ and $\omega_2$ are orthogonal, we get that $KLK’L’$ is harmonic, hence $MK$ and $ML$ are tangent to $(KLK’L’)$ . Hence we are done.

This post has been edited 1 time. Last edited by Ywgh1, Aug 13, 2024, 8:52 AM

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Eka01

204 posts

Eka01

#127 h Aug 16, 2024, 5:11 PM • 2 Y

Y by Sammy27, AaruPhyMath

Let $BX$ and $AX$ meet $(A,AC)$ and $(B,BC)$ again at $V$ and $U$ respectively.
It is obvious that the above circles are orthogonal and that $X$ lies on the radical axis of these two circles, hence by converse of power of point
$(KLUV)$ is cyclic. Also $AC^2=AK.AU=AL^2$ so $AL$ and similarly $BK$ is tangent to $(KLUV)$ so $M$ is the pole of $KL$ which implies what we needed to show.

Attachments:

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dolphinday

1310 posts

dolphinday

#128 h Aug 25, 2024, 6:54 PM

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Let $\omega_A$ be the circle with center $A$ passing through $C$ and define $\omega_B$ similarly.
Let the second intersections of $AK$ and $BL$ with $\omega_B$ and $\omega_A$ be $M$ and $N$ respectively.
Since $\angle ACB = 90^\circ$ we have that $AC$ and $BC$ are tangents to $\omega_B$ and $\omega_A$ respectively. Then by Radical Axis since $KM$ and $LN$ intersect on the radical axis of $\omega_A$ and $\omega_B$ , $KLMN$ is cyclic.
Since $\omega_A$ and $\omega_B$ are orthogonal, $K$ and $M$ are inverses wrt $\omega_A \implies AK \cdot AM = AL^2$ so $AL$ is a tangent to $(KLMN)$ and $BK$ is as well. So this implies that $MK$ and $ML$ are tangents so $MK = ML$ , done.

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KevinYang2.71

393 posts

KevinYang2.71

#129 h Today at 2:51 AM

Y by

We use barycentric coordinates with $A=(1,\,0,\,0),\,B=(0,\,1,\,0),\,C=(0,\,0,\,1)$ . Note that $a^2+b^2=c^2$ and $D=(a^2:b^2:0)$ . Let $X=:(a^2:b^2:t)$ and suppose $K=(k:b^2:t)=\left(\frac{k}{k+b^2+t},\,\frac{b^2}{k+b^2+t},\,\frac{t}{k+b^2+t}\right)$ . Then $\overrightarrow{BK}=\left(\frac{k}{k+b^2+t},\,-\frac{k+t}{k+b^2+t},\,\frac{t}{k+b^2+t}\right)$ , so from $BK=BC$ we get
$\begin{align*} a^2&=\frac{a^2t(k+t)-b^2kt+c^2k(k+t)}{(k+b^2+t)^2}\\ &=\frac{a^2t(k+t)-b^2kt+(a^2+b^2)k(k+t)}{(k+b^2+t)^2}\\ &=\frac{a^2(k+t)^2+b^2k}{(k+b^2+t)^2}\\ \implies b^2k^2&=a^2((k+b^2+t)^2-(k+t)^2)\\ &=a^2b^2(b^2+2k+2t). \end{align*}$ Hence $k^2-2a^2k-a^2(b^2+2t)=0$ so
$k=\frac{2a^2+\sqrt{4a^4+4a^2(b^2+2t)}}{2}=a^2+a\sqrt{c^2+2t}.$ Similarly, $L=\left(a^2:b^2+b\sqrt{c^2+2t}:t\right)$ . Now
$M=\left(a^2+a\sqrt{c^2+2t}:b^2+b\sqrt{c^2+2t}:t\right)=\left(\frac{a^2+a\sqrt{c^2+2t}}{\alpha},\,\frac{b^2+b\sqrt{c^2+2t}}{\alpha},\,\frac{t}{\alpha}\right),$ where $\alpha:=c^2+t+(a+b)\sqrt{c^2+2t}$ . Let $\beta:=k+b^2+t=c^2+t+a\sqrt{c^2+2t}$ so
$\overrightarrow{KM}=\left(\left(a^2+a\sqrt{c+2t}\right)\left(\frac{1}{\alpha}-\frac{1}{\beta}\right),\,b^2\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)+\frac{b\sqrt{c+2t}}{\alpha},\,t\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)\right).$ Noting that $\frac{1}{\alpha}-\frac{1}{\beta}=-\frac{b\sqrt{c^2+2t}}{\alpha\beta}$ , we get
$\overrightarrow{KM}=\frac{b\sqrt{c^2+2t}}{\alpha\beta}\left(-\left(a^2+a\sqrt{c+2t}\right),\,\beta-b^2,\,-t\right)$ Note that $\frac{\sqrt{c^2+2t}}{\alpha}$ is independent of $a$ and $b$ . Now
$\begin{align*} \left(\frac{\alpha}{\sqrt{c^2+2t}}KM\right)^2&=\frac{b^2}{\beta^2}\left(a^2t(\beta-b^2)-b^2t\left(a^2+a\sqrt{c^2+2t}\right)+c^2(\beta-b^2)\left(a^2+a\sqrt{c^2+2t}\right)\right)\\ &=\frac{ab^2}{\beta^2}\left((\beta-b^2)\left(at+ac^2+c^2\sqrt{c^2+2t}\right)-b^2t\left(a+\sqrt{c^2+2t}\right)\right)\\ &=\frac{ab^2}{\beta^2}\left((\beta-b^2)\left(a\beta+b^2\sqrt{c^2+2t}\right)-b^2t\left(a+\sqrt{c^2+2t}\right)\right)\\ &=\frac{ab^2}{\beta^2}\left(a\beta(\beta-b^2)+b^2\sqrt{c^2+2t}\left(a^2+t+a\sqrt{c^2+2t}\right)-b^2t\left(a+\sqrt{c^2+2t}\right)\right)\\ &=\frac{a^2b^2}{\beta^2}\left(\beta(\beta-b^2)+b^2\sqrt{c^2+2t}\left(a+\sqrt{c^2+2t}\right)-b^2t\right)\\ &=\frac{a^2b^2}{\beta^2}\left(\beta^2+b^2\left(\sqrt{c^2+2t}\left(a+\sqrt{c^2+2t}\right)-t-\beta\right)\right)\\ &=\frac{a^2b^2}{\beta^2}\left(\beta^2+b^2\left(a\sqrt{c^2+2t}+c^2+2t-t-\beta\right)\right)\\ &=a^2b^2, \end{align*}$ which is independent of $a$ and $b$ . Thus $KM=LM$ . $\square$

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