RMM 2019 Problem 2

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math90

1474 posts

math90

#1 h Feb 23, 2019, 11:40 AM • 6 Y

Y by adityaguharoy, EthanTAoPS, mathematicsy, Adventure10, Mango247, eggymath

Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$ . Let $E$ be the midpoint of $AC$ . Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$ , respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$ . Prove that $PE$ is tangent to $\Omega$ .

Jakob Jurij Snoj, Slovenia

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Reason: Thanks #13

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rmtf1111

698 posts

rmtf1111

#2 h Feb 23, 2019, 11:46 AM • 2 Y

Y by Adventure10, Mango247

Suppose that $ABCD$ is not a rectangle and let $F$ be the midpoint of $BD$ . Let $l_1$ and $l_2$ be the tangents to $\Omega$ at $E$ and $D$ respectively and let $m_1$ and $m_2$ be the tangents to $\omega$ at $A$ and $F$ respectively. Suppose that $m_2$ intersect $\overline{CD}$ and $\overline{AB}$ at $V$ and $V_1$ respetively and let $U$ be on $\overline{CD}$ such that $UF$ is tangent to $\Omega$ . Due to the tangency relation, we have that $\angle{CUF}=\angle{ECF}$ and $\angle{AV_1F}=\angle{EAF}$ , thus by sine theorem and similarity the following relations hold $\frac{UD}{DF}=\frac{UF}{CF}=\frac{\sin{\angle{CFE}}}{\sin{\angle{ECF}}}=\frac{\sin{\angle{AFE}}}{\sin{\angle{CAF}}}=\frac{\sin{\angle{FAV_1}}}{\sin{\angle{AV_1F}}}=\frac{FV_1}{AF}=\frac{BV_1}{BF} \implies BV_1=DV=DU$ This clearly implies that the pencil $(FD,FE;m_2,FU)$ is harmonic, thus $m_2,l_1$ and $l_2$ are concurrent. Analogously $l_1,m_1$ and $m_2$ are also concurrent, thus $G$ lies on both $m_2$ and $m_1$ .

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TheDarkPrince

3042 posts

TheDarkPrince

#3 h Feb 23, 2019, 11:56 AM • 1 Y

Y by Adventure10

Solution. Let $F$ be the midpoint of $BD$ . Let $P'$ be the intersection of tangents from $E$ and $D$ to $\odot(DCE)$ . We'll prove that $P'F$ is tangent to $\odot(ABE)$ which would give from symmetry that $P'A$ is tangent to $\odot (ABE)$ and we would be done.

Now, invert about $E$ with arbitrary radius.

Rephrased problem wrote:

Let $FAC$ be a triangle and $E$ be the midpoint of $AC$ . Let $D$ be a point on $CF$ such that $EF$ is tangent to $\odot(DCE)$ and $B$ is a point on $AF$ such that $EF$ is tangent to $\odot(BEA)$ . Let $P'$ be a point such that $EP'||DF$ and $DF$ is tangent to $\odot(P'ED)$ . Prove that $\odot(P'EF)$ is tangent to $AF$ .

(Proof): Note that $DP' = DE$ and $\triangle DEF \sim \triangle ECF$ so $\frac{AE}{EF}=\frac{CE}{EF} = \frac{DE}{DF}=\frac{DP'}{DF}.$ Also, $\angle P'DF = \angle ECF + \angle CFE = \angle AEF.$ Therefore $\triangle AEF \sim \triangle P'DF$ so $\angle AFE = \angle DFP' = \angle EP'F$ and we are done. $~\square$

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Math1Zzang

62 posts

Math1Zzang

#4 h Feb 23, 2019, 12:06 PM • 4 Y

Y by the_99999th_user, Adventure10, Mango247, Mango247

Let $F$ be the midpoint of $BD$ . Easily $DF=CE=AE$ . Let $P'$ be the intersection of bisector of $AF$ and $DE$ . Then $\triangle P'DF \equiv \triangle P'EA(SSS)$ . Let $S$ be the intersection of $AC$ and $BD$ . After some angle chasing, $P'S\parallel AB$ . After more angle chasing, $P'D$ , $P'E$ are tangent to $\Omega$ and $P'A$ , $P'F$ are tangent to $\omega$ . So $P'=P$ and $PE$ is tangent to $\Omega$ .

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v_Enhance

6862 posts

v_Enhance

#5 h Feb 23, 2019, 1:07 PM • 10 Y

Y by richrow12, Michcio, e_plus_pi, RAMUGAUSS, Kayak, v4913, HamstPan38825, mathematicsy, ChanandlerBong, Adventure10

Let $F$ denote the midpoint of $\overline{BD}$ , so $CDFE$ and $ABFE$ are cyclic trapezoids. We use complex numbers with respect to $CDEF$ .

$[asy] pair C = dir(130); pair E = dir(210); pair D = dir(330); pair F = C*D/E; pair A = 2*E-C; pair B = 2*F-D; filldraw(unitcircle, invisible, deepcyan); draw(A--B--C--D--cycle, lightred); draw(E--F, orange); draw(B--A, orange); draw(C--D, orange); draw(C--A, lightred); draw(B--D, lightred); pair P = 2*E*D/(E+D); draw(E--P--D, deepcyan); draw(E--D, deepcyan); draw(P--A, deepcyan); draw(circumcircle(E, F, A), dotted+deepgreen); pair O = origin; pair M = midpoint(E--D); draw(O--P, dotted+blue); dot("$C$", C, dir(C)); dot("$E$", E, dir(E)); dot("$D$", D, dir(D)); dot("$F$", F, dir(F)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$P$", P, dir(P)); dot("$O$", O, dir(10)); dot(M); /* TSQ Source: C = dir 130 E = dir 210 D = dir 330 F = C*D/E A = 2*E-C B = 2*F-D unitcircle 0.1 lightcyan / deepcyan A--B--C--D--cycle lightred E--F orange B--A orange C--D orange C--A lightred B--D lightred P = 2*E*D/(E+D) E--P--D deepcyan E--D deepcyan P--A deepcyan circumcircle E F A dotted deepgreen O = origin R10 M .= midpoint E--D R315 O--P dotted blue */ [/asy]$

Let $C = x$ , $E = y$ , $D = z$ , so $F = xz/y$ . Re-define $P = \frac{2yz}{y+z}$ as the intersection of the tangents to $\Omega$ at $E$ and $D$ ; then we wish to show $\overline{PA}$ is tangent to $\Gamma$ . Compute $\frac{F-A}{P-A} = \frac{\frac{xz}{y} - (2y-x)}{\frac{2yz}{y+z} - (2y-x)} = \frac{y+z}{y} \frac{2xz-y(2y-x)}{2yz-(y+z)(2y-x)} = \frac{y+z}{y}.$ Hence $\measuredangle FAP$ equals $\measuredangle EOP = \measuredangle ECD = \measuredangle ACD = \measuredangle AEF$ , where $O$ is the center of $\Omega$ . This implies the desired tangency.

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MathInfinite

187 posts

MathInfinite

#6 h Feb 23, 2019, 1:15 PM • 7 Y

Y by rmtf1111, Steff9, ILOVEMYFAMILY, Adventure10, Mango247, Mango247, Mango247

Here’s my solution.

Let $F$ denote the midpoint of $BD$ and let $X = AC \cap BD$ .
Redefine $P$ as the intersection of perpendicular bisectors of $AF, DE$ .

Note that $\Delta PDF \equiv \Delta PEA$ as $PE = PD, PF = PA, DF = DA$ .
I claim that

$P,X,E,D$ are concyclic

This is easy to see since $\angle{PDF} = \angle{PDX} = \angle{PEA} = \angle{PEX}$ .
So, $\angle{PDE} = 180^{\circ}-\frac{\angle{DPE}}{2} = 180^{\circ}-\frac{\angle{DXE}}{2} = \angle{ECD}$ .
Thus, $DP$ is tangent to $(CED)$ . Similarly, $AP$ is tangent to $(ABE)$ .
Since, $DP = EP$ , we have $EP$ tangent to $(CED)$ . $\blacksquare$ .

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Reason: There’s a shorter solution

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Crimson.

8 posts

Crimson.

#7 h Feb 23, 2019, 1:16 PM • 2 Y

Y by pieater314159, Adventure10

This problem can be easily solved with complex numbers:
Let $F$ be the reflection of $E$ in the perpendicular bisector of $AB$ .Hence $ABFE$ and $CDEF$ are cyclic trapezoids.Now set the circle $(CDEF)$ as unit circle and let $P'$ be the intersection of tangents at $D$ and $E$ to $(CEF)$ .We want to prove that $AP'$ is tangent to $(AFE)$ .Denote by $O$ the circumcenter of $(AFE)$ .We just have to prove $OA\perp AP'$
$p'=\frac{2de}{d+e}$ and $e-a=c-e$ and as $FE\parallel CD$ we have $c=\frac{ef}{d}$ so $a=e\cdot\frac{2d-f}{d}$
$o-a=\frac{(f-a)(e-a)(\overline{f-e})}{(\overline{f-a})(e-a)-(\overline{e-a})(f-a)}$
$\frac{o-a}{\overline{o-a}}=\frac{f-a}{\overline{f-a}}\cdot\frac{c-e}{\overline{c-e}}\cdot\frac{\overline{f-e}}{f-e}\cdot\frac{-(\overline{f-a})(e-a)+(\overline{e-a})(f-a)}{(\overline{f-a})(e-a)-(\overline{e-a})(f-a)}$ So we have $\frac{o-a}{\overline{o-a}}=\frac{f-a}{\overline{f-a}}\cdot(-ce)\cdot(-\frac{1}{ef})\cdot(-1)=-\frac{f-a}{\overline{f-a}}\cdot\frac{e}{d}$ Now $f-a=f-(e\cdot\frac{2d-f}{d})=\frac{df+fe-2de}{d}$ and $a-p'=a-\frac{2de}{d+e}=e\bigg(\frac{2d-f}{d}-\frac{2d}{d+e}\bigg)=e\cdot\frac{2de-ef-fd}{d(d+e)}$
$\frac{a-p'}{\overline{a-p'}}=\frac{e}{d+e}\cdot \frac{1}{\frac{1}{e}}\cdot\frac{d+e}{de}\cdot\frac{f-a}{\overline{f-a}}=-\frac{o-a}{\overline{o-a}}$ Hence we are done.

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math90

1474 posts

math90

#8 h Feb 23, 2019, 1:52 PM • 1 Y

Y by Adventure10

Is there any geometric solution without redefining $P$ ?

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pslove

72 posts

pslove

#9 h Feb 23, 2019, 2:36 PM • 3 Y

Y by math90, Adventure10, Mango247

math90 wrote:

Is there any geometric solution without redefining $P$ ?

The second official solution is a geometric solution without redefining $P$ . It uses isogonal conjugates.

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MNJ2357

644 posts

MNJ2357

#10 h Feb 23, 2019, 2:40 PM • 2 Y

Y by Adventure10, Mango247

Am I the only one who didn't use the midpoint of $BD$ ? :maybe:

Let $Q$ be the point such that $\triangle QDA\equiv\triangle EBC$ , and suppose $DQ$ meet $AB$ at $T$ . Let $M,N$ be the midpoints of $BC,AD$ . (Also let $P'$ be the point such that $P'E$ and $P'D$ are both tangent to $\Omega$ . It suffices to prove that $PA$ is tangent to $(ABE)$ .) One can easily prove that $K,N,E,M$ are collinear. Since $QE\parallel AB$ , and note that $AQ=CE=AE$ , so
$\angle QEA=\angle DCA=\angle PED\Longrightarrow \triangle QEA\sim \triangle DEP$ . We that $\triangle QED\sim\triangle AEP$ by spiral similarity, Thus
$\angle PAE=\angle DQE=\angle DTC$ , but since $\frac{AT+CD}{2}=QE=\frac{AB+CD}{2}\Longrightarrow \angle DTC=\angle EBC$ , so we're done. ( $\square BTQE$ is a cyclic trapezoid since $Q$ and $E$ are symmetric with respect to the perpendicular bisector of $BT$ .)

Edit : Thanks @below

This post has been edited 2 times. Last edited by MNJ2357, Feb 23, 2019, 3:05 PM
Reason: P->E

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pslove

72 posts

pslove

#11 h Feb 23, 2019, 3:04 PM • 1 Y

Y by Adventure10

MNJ2357 wrote:

Let $Q$ be the point such that $\triangle QDA\equiv\triangle PBC$

I think you mean $\triangle QDA\equiv \triangle EBC$ ?

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richrow12

411 posts

richrow12

#12 h Feb 23, 2019, 3:04 PM • 1 Y

Y by Adventure10

Let's $P_1$ be the intersection of tangents to $(AEFB)$ at points $A$ and $F$ and let's $P_2$ be the intersection of tangents to $(CFED)$ at points $C$ and $E$ . We will prove that $P_1=P_2$ . Denote $K$ as the interesection of lines $AC$ and $BD$ . Firstly, note that triangles $AP_1F$ and $EKF$ are similar (since $\angle KEF=\angle PAF$ and $\angle KFE=\angle PFA$ ). Therefore, $\angle AP_1F=\angle AKF$ , so $AP_1KF$ is cyclic. From $AP=PF$ we obtain that $P_1$ lies on the angle bissector of $\angle AKD$ . Also, siimilarity $AP_1F$ and $FKE$ implies similarity $AEF$ and $P_1KF$ . Hence,
$\frac{KP_1}{KF}=\frac{AE}{EF}.$ Analagously, triangles $DP_2E$ and $EKF$ are similar, $P_2$ lies on the angle bissector of $\angle AKD$ . Also, $P_2EK$ similar to $DEF$ , so
$\frac{KP_2}{KE}=\frac{FD}{EF}.$ Now, note that
$KP_1=\frac{AE\cdot KF}{EF}=\frac{FD\cdot KE}{EF}=KP_2.$ But points $P_1$ and $P_2$ lie on the angle bissector $\angle AKD$ , so $P_1=P_2$ . Therefore, $P=P_1=P_2$ , as desired.

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L3435

104 posts

L3435

#13 h Feb 23, 2019, 3:06 PM • 1 Y

Y by Adventure10

math90 wrote:

Let $ABCD$ be an isosceles trapezoid with $AB\parallel CD$ . Let $E$ be the midpoint of $AC$ . Denote by $\omega$ and $\Omega$ the circumcircles of the triangles $ABE$ and $CDE$ , respectively. Let $P$ be the crossing point of the tangent to $\omega$ at $A$ with the tangent to $\Omega$ at $D$ . Prove that $PE$ is tangent to $\Omega$ .

Jakob Jurij Snoj, Slovenia

FTFY

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ABCDE

1963 posts

ABCDE

#14 h Feb 23, 2019, 3:50 PM • 5 Y

Y by anantmudgal09, tenplusten, nikitadas, Adventure10, Mango247

Let $F$ be the midpoint of $BD$ and $G=AF\cap CD,H=AB\cap DE$ . Clearly, $F\in(ABE),(CDE)$ . By 2019 USA TST P1, the tangents to $(FGD),(ECD),(AEF)$ at $F,E,A$ concur and the tangents to $(FBA),(EHA),(DEF)$ at $F,E,D$ concur. But $(FGD),(ABEF)$ and $(EHA),(CDEF)$ are tangent, so the problem follows.

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Pathological

578 posts

Pathological

#15 h Feb 23, 2019, 3:58 PM • 3 Y

Y by JasperL, Adventure10, Mango247

Another inversion solution

Solution

This post has been edited 1 time. Last edited by Pathological, Aug 19, 2019, 8:06 PM

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Martin2001

126 posts

Martin2001

#69 h Jan 6, 2024, 3:36 AM

Y by

Alright, all aboard the crappy variable name train. We will complex bash this problem. Let $\Omega$ be the unit circle, and let $F$ be the second intersection of $\Omega \cap \Gamma.$
We let $C=a, E=b, D=c,$ so $F=\frac{ab}{c}.$ We prove that $P=\frac{2ab}{a+b},$ which is the intersection point from the tangents at $D$ and $E,$ is on $\overline{AA}.$ This means that we need $\angle FAP= \angle AEF$ for our claim to be true.
Note that
$\frac{F-A}{P-A}=\frac{\frac{xz}{y}-(2y-x)}{\frac{2yz}{y+z}-(2y-x)}=\frac{y+z}{y}.$ Note that $\frac{D}{P}=\frac{z}{\frac{2yz}{y+z}}=\frac{y+z}{2y},$ so $\angle FAP=\angle DOP=\angle ECD=\angle ACD,$ which $\angle ACD=\angle AEF$ becaues $\overline{EF} \parallel \overline{DC},$ and as there is only one point $\overline{AA} \cap \overline{DD},$ this point is also on $\overline{EE},$ and we're done.

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Scilyse

386 posts

Scilyse

#70 h Feb 9, 2024, 7:41 AM

Y by

We use phantom points (and $e$ as a free variable). Define $P'$ as the intersection of the tangents to $\Omega$ at $D$ and $E$ . Additionally, set $\Omega$ as the unit circle, and let $d = \overline{c}$ , $b = \overline{a}$ . We wish to prove $P = P'$ , or equivalently that $P'A$ is tangent to $\Gamma$ , and we do this by proving $\angle P'AE = \angle ABE$ .

Note
$\begin{align*} \frac{e - a}{p' - a} \div \frac{e - b}{a - b} &= \frac{(e - a)(a - b)}{(p' - a)(e - b)} \\ &= \frac{(e - (2e - c))(2e - c - \overline{2e - c})}{\left(\frac{2de}{d + e} - (2e - c)\right)(e - \overline{2e - c})} \\ &= \frac{(c - e)(2e - c - \overline{2e - c})}{\left(\frac{2\overline{c} e}{\overline{c} + e} + c - 2e\right)\left(e - \frac 2e + \frac 1c\right)} \\ &= \frac{(c - e)(2e - c - \overline{2e - c})}{\left(\frac{2e}{1 + ce} + c - 2e\right)\left(\frac{ce^2 - 2c + e}{ce}\right)} \\ &= \frac{ce(1 + ce)(c - e)(2e - c - \overline{2e - c})}{(2e + (c - 2e)(1 + ce))(ce^2 - 2c + e)} \\ &= \frac{ce(1 + ce)(c - e)(2e - c - \overline{2e - c})}{(c + c^2 e - 2ce^2)(ce^2 - 2c + e)} \\ \overline{\frac{e - a}{p' - a} \div \frac{e - b}{a - b}} &= \frac{\frac{1}{ce} \left(\frac{1 + ce}{ce}\right) \left(\frac{e - c}{ce}\right) (\overline{2e - c} - (2e - c))}{\left(\frac 1c + \frac{1}{c^2 e} - \frac{2}{ce^2}\right)\left(\frac{1}{ce^2} - \frac 2c + \frac 1e\right)} \\ &= \frac{ce(1 + ce)(e - c)(\overline{2e - c} - (2e - c))}{c^4 e^4 \left(\frac 1c + \frac{1}{c^2 e} - \frac{2}{ce^2}\right)\left(\frac{1}{ce^2} - \frac 2c + \frac 1e\right)} \\ &= \frac{ce(1 + ce)(c - e)(2e - c - \overline{2e - c})}{(ce^2 + e - 2c)(c - 2ce^2 + c^2 e)} \\ &= \frac{e - a}{p' - a} \div \frac{e - b}{a - b}\text{.} \end{align*}$
Therefore $\frac{e - a}{p' - a} \div \frac{e - b}{a - b} \in \mathbb{R}$ and we are done by alternate segment theorem.

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StefanSebez

53 posts

StefanSebez

#71 h Feb 15, 2024, 10:54 PM

Y by

Seems like this is a new solution

Redefine $P$ such that $PE, PD$ are tangent to $(CDE)$
Draw point $F\in (ABCD)$ such that $AD=AF$
Both $\Delta AFD$ and $\Delta PED$ are isosceles and $\angle DFA=\angle DCA=\angle DEP$
$\implies \Delta AFD \sim \Delta PED$
So a spiral similarity at $D$ takes $\Delta AFD$ to $\Delta PED$ , so we get $\Delta DPA \sim DEF$
$AF=AD=BC$ so $ACBF$ is isosceles trapezoid $\implies \angle AEF=\angle BEC$ (because $EA=EC$ )
Now we can angle chase the rest
Let $\angle BAC=\alpha, \angle CBA=\beta, \angle EDC=\theta, \angle EBA=\phi$
$\angle DPA=\angle DEF=\angle DEA+\angle AEF=\theta+\alpha+\angle BEC=\theta+\phi+2\alpha$
$\angle EAP=\angle EAD-\angle PAD=(\beta-\alpha)-(180-\angle ADP-\angle DPA) \newline =\beta-\alpha-180+(180-\beta-\alpha-\theta)+(\theta+\phi+2\alpha)=\phi$
So $PA$ is tangent to $(AEB)$ as desired

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GrantStar

813 posts

GrantStar

#72 h Feb 16, 2024, 3:54 AM • 1 Y

Y by OronSH

Let $F$ be the midpoint of $BC$ , on both circles. Let $G$ be the point on $\Omega$ such that $(GF;ED)=-1$ . Let $GF$ hit $ED$ at $J$ .
Redefine $P$ as the intersections of the tangents at $E,D$ to $\Omega$ . Thus $P,F,J,G$ are collinear.

Claim: $FP$ is tangent to $\omega$ .
Proof. Let $G'=CG\cap EF$ . As $-1=(GF;ED)\overset{C}{=}(G'F;E\infty)$ so $E$ is the midpoint of $G'F$ . Thus $AFCG'$ is a parallelogram, so $CG \parallel AF$ . Reim finishes. $\blacksquare$

Let $PE$ intersect $AF$ at $I$ and $\omega$ at $H$ .

Claim: $EFIJ$ is cyclic
Proof. The proof is just angles. $\measuredangle JFI=\measuredangle PFI=\measuredangle FBA=\measuredangle FDC =\measuredangle FDE+\measuredangle EDC=\measuredangle FEI+\measuredangle JEF=\measuredangle JEI$ $\blacksquare$
Therefore, by reims on $\omega$ and $\Omega$ , $FH\parallel EG$ . Similarly, reims on $(EFJI)$ and $\omega$ gives $IJ\parallel FH$ . Thus, $-1=(PJ;FG)\overset{\infty_{FH}}{=}(PI;HE)\overset{F}{=}(FA;HE)$ Hence $EH$ is a symmedian and thus $P$ is the intersection of the tangents at $A$ and $F$ to $\omega$ . We thus conclude the result.
Remark: Very good.

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Leo.Euler

577 posts

Leo.Euler

#73 h Feb 20, 2024, 7:42 AM

Y by

Let $K = \overline{AC} \cap \overline{BD}$ and $F$ denote the midpoint of $BD$ .

Claim: Let $Q$ be the second intersection of $(KAF)$ and $(KED)$ . Then $\overline{QK} \parallel \overline{CD}$ .
Proof. Let $f(\bullet) = \text{Pow}(\bullet, (KAF)) - \text{Pow}(\bullet, (KED))$ . Then it suffices to show that $f(C)=f(D)$ . Compute $f(C)-f(D) = (CK \cdot CA - CE \cdot CK) - (DF \cdot DK - 0) = CK \cdot EA - DF \cdot DK= CK \cdot EC - DF \cdot DK = 0,$ as desired.
:yoda:

Claim: In fact, $P=Q$ .
Proof. Note that as $ABEF$ is an isosceles trapezoid, the center of $(ABEF)$ lies on $(KAF)$ . Thus, $P$ lies on $(KAF)$ . I contend that $\overline{PK} \parallel \overline{FE}$ , which would imply $P=Q$ by the prior claim. By simple angle chasing we have $\angle PKA = \angle PFA = \angle FEA,$ which implies the desired parallelism.
:yoda:

Claim: Finally, $PD=PE$ .
Proof. Note that $P$ is the center of the spiral similarity mapping $\overline{AF} \mapsto \overline{ED}$ . Thus, $\triangle FPA \sim \triangle DPE$ , so as $FP=AP$ , we have $DP=EP$ , as desired.
:yoda:

We conclude from the final claim.

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Shreyasharma

666 posts

Shreyasharma

#74 h Feb 20, 2024, 9:15 AM

Y by

Set $(DEC)$ as the unit circle and redefine $P$ as the point such that $\overline{PD}$ and $\overline{PE}$ are tangent to $\Omega$ . Orient so that $\overline{CD}$ is perpendicular to the real axis. Define $F$ as the midpoint of $\overline{BD}$ . Compute,
$\begin{align*} d &= 1/c\\ f &= 1/e\\ a &= 2e - c\\ p &= \frac{2de}{d+e} = \frac{2e/c}{1/c + e} = \frac{2e}{ce + 1} \end{align*}$ However clearly we have,
$\begin{align*} \frac{p - a}{f - a} &= \frac{\frac{2e - (2e - c)(ce + 1)}{ce + 1}}{1/e - 2e + c}\\ &= \frac{\frac{2e - 2e - 2e^2c + c^2e + c}{ce + 1}}{1/e(1 - 2e^2 + ce)}\\ &= \frac{\frac{c(2e^2 + ce + 1)}{ce + 1}}{1/e(1 - 2e^2 + ce)}\\ &= \frac{ce}{ce + 1} \end{align*}$ Also we have,
$\begin{align*} \frac{a - c}{d - c} &= \frac{2e - 2c}{1/c - c}\\ &= \frac{2c(e - c)}{1 - c^2} \end{align*}$ Then dividing we have,
$\begin{align*} \frac{2(e-c)(ec + 1)}{e(1 - c^2)} \end{align*}$ Taking the conjugate this is simply,
$\begin{align*} \frac{2/c^2e^2(c - e)(ec + 1)}{1/c^2e(c^2 - 1)} = \frac{2(e - c)(ec+1)}{e(1-c^2)} \end{align*}$ and so we find $\angle PAF = \angle DCA = \angle FEA$ proving the tangency.

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Thapakazi

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Thapakazi

#75 h Mar 28, 2024, 1:55 PM

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Let $F$ be the other intersection of $(ABC)$ and $(CDE)$ , redefine $P$ to be the intersection of tangents through $D$ and $E$ of $(DEC).$ Then, $B,F,E$ are collinear as:

$\measuredangle BFE = \measuredangle BAE = \measuredangle ACD = -\measuredangle EFD.$
Furthermore, this also implies $EF \parallel BC$ as

$\measuredangle BFE = \measuredangle BAC = \measuredangle BDC = \measuredangle FDC.$
Now, we use complex numbers with $(DFEC)$ as the unit circle.

We see that $p = \frac{2de}{d+e}$ , $f = cd/e$ , and $a = 2e - c.$ We compute

$\begin{align*} \frac{a-p}{a-f} \div \frac{e-a}{e-f} &= \frac{2e-c-\frac{2de}{d+e}}{2e-c-\frac{cd}{e}} \times \frac{e-\frac{cd}{e}}{c-e} \\ &= \frac{e}{d+e} \times \frac{(d+e)(2e-c)-2de}{2e^2-ce-cd} \times \frac{e^2-cd}{e(c-e)} \\ &= \frac{1}{d+e} \times \frac{2e^2 - cd - ce}{2e^2 - ce - cd} \times \frac{e^2-cd}{c-e}\\ &= \frac{e^2-cd}{(d+e)(c-e).} \end{align*}$
This has conjugate

$\frac{\frac{1}{e^2} - \frac{1}{cd}}{(\frac{1}{d} + \frac{1}{e})(\frac{1}{c}-\frac{1}{e})} = \frac{cd-e^2}{(e+d)(e-d)} = \frac{a-p}{a-f} \div \frac{e-a}{e-f}.$
Hence, $\frac{a-p}{a-f} \div \frac{e-a}{e-f} \in \mathbb{R}$ giving us $\measuredangle FAP = \measuredangle FEA$ as needed.

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cursed_tangent1434

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cursed_tangent1434

#76 h Mar 28, 2024, 2:39 PM

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Set $\Omega$ be the unit circle. WLOG, rotate so that $CD$ is parallel to the real axis. Then, $d=-\frac{1}{c}$ . We can then obtain,
$a=2e-c$ and,
$b=-\overline{a}=-\overline{2e-c}=\frac{e-2c}{ce}$ Let $P'$ (denoted by the complex number $p$ ) be the intersection of the tangents to $\Omega$ at $D$ and $E$ .Then, we can also easily obtain that,
$\begin{align*} p &= \frac{2de}{d+e}\\ &= \frac{2\left(-\frac{1}{c}\right)e}{e-\frac{1}{c}}\\ &= \frac{-\frac{2e}{c}}{\frac{ce-1}{c}}\\ &= \frac{2e}{1-ec} \end{align*}$ Now, we wish to show $\measuredangle PAE = \measuredangle ABE$ . Thus, we require,
$\arg \left(\frac{e-a}{p-a}\right)=\arg\left(\frac{e-b}{a-b}\right)$ Thus, we must have $A = \frac{(e-a)(a-b)}{(p-a)(e-b)} \in \mathbb{R}$ . Note that,
$\begin{align*} A &= \frac{(e-a)(a-b)}{(p-a)(e-b)}\\ &= \frac{(e-2e+c)(2e-c+\frac{2c-e}{ce})}{\left( \frac{2e}{1-ce}-2e+c \right)\left( e+\frac{2c-e}{ce} \right)}\\ &= \frac{(c-e)(1-ce)(2ce^2-c^2e+2c-e)}{(2ce^2-c^2e+c)(2c+ce^2-e)} \end{align*}$ Now, we also have that,
$\begin{align*} \overline{A} &= \overline{\left(\frac{(c-e)(1-ce)(2ce^2-c^2e+2c-e)}{(2ce^2-c^2e+c)(2c+ce^2-e)}\right)}\\ &= \frac{\left(\frac{1}{c}-\frac{1}{e}\right)\left(\frac{2}{ce^2}-\frac{1}{c^2e}+\frac{2}{c}-\frac{1}{e}\right)\left(1-\frac{1}{ce}\right)}{\left( \frac{2}{ce^2}-\frac{1}{c^2e}+\frac{1}{c} \right)\left(\frac{2}{c}+\frac{1}{ce^2}-\frac{1}{e}\right)}\\ &= \frac{\frac{(e-c)}{ec}\left( \frac{2c-e+2ce^2-c^2e}{c^2e^2} \right)\left(\frac{ce-1}{ce}\right)}{\left(\frac{2c-e+ce^2}{c^2e^2}\right)\left( \frac{2e^2-ce+1}{ce^2} \right)}\\ &= \frac{c^3e^4(c-e)(1-ec)(2ce^2-c^2e+2c-e)}{c^4e^4(ce^2+2c-e)(2e^2+1-ec)}\\ &=\frac{(c-e)(1-ce)(2ce^2-c^2e+2c-e)}{(2ce^2-c^2e+c)(2c+ce^2-e)} \end{align*}$ Thus, $\overline{A}=A$ and indeed, $\frac{(e-a)(a-b)}{(p-a)(e-b)} \in \mathbb{R}$ which implies that $\measuredangle PAE = \measuredangle ABE$ . Thus, $\overline{PA}$ is tangent to $\Gamma$ at $A$ implying that $P'=P$ . Thus, indeed $\overline{PE}$ is tangent to $\Omega$ as claimed.

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OronSH

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OronSH

#77 h May 6, 2024, 4:49 PM

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Let $F$ be the midpoint of $BD.$ Redefine $P$ as the center of spiral similarity sending $DF$ to $EA.$ Since $DF=EA$ we have $\triangle DPF\cong\triangle EPA$ so $DP=EP,FP=AP.$ Let $X$ be the intersection of $DF$ and $EA.$ Then we have $PXED$ cyclic so $\measuredangle DPE=\measuredangle DXE=2\measuredangle DCA,$ but this implies $PD,PE$ are tangent to $(CDFE).$ Similarly we get $PA,PF$ are tangent to $(ABEF),$ so we are done.

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Ywgh1

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Ywgh1

#78 h Jun 9, 2024, 2:54 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.521269841269856cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 1.253354497354498, xmax = 14.774624338624355, ymin = -4.519809523809521, ymax = 1.9744973544973539; /* image dimensions */ pen qqzzcc = rgb(0.,0.6,0.8); pen ccccff = rgb(0.8,0.8,1.); pen qqccqq = rgb(0.,0.8,0.); pen ttffcc = rgb(0.2,1.,0.8); pen zzwwff = rgb(0.6,0.4,1.); pen zzffqq = rgb(0.6,1.,0.); pen ccwwff = rgb(0.8,0.4,1.); pen ffqqff = rgb(1.,0.,1.); pen afeeee = rgb(0.6862745098039216,0.9333333333333333,0.9333333333333333); pen ffdxqq = rgb(1.,0.8431372549019608,0.); pen ffcqcb = rgb(1.,0.7529411764705882,0.796078431372549); draw((4.374406723856454,-0.2667693776957385)--(5.536649473250011,-1.6410921261999398)--(2.94,-4.24)--cycle, linewidth(1.) + ffcqcb); draw((4.374406723856454,-0.2667693776957385)--(8.380754882891011,-1.6058954569210449)--(5.720585931236688,0.9279576903792971)--cycle, linewidth(1.) + ffcqcb); /* draw figures */ draw((5.720585931236688,0.9279576903792971)--(8.133298946500023,0.9578157476001203), linewidth(1.) + qqzzcc); draw((2.94,-4.24)--(11.040923834545335,-4.139748604221387), linewidth(1.) + qqzzcc); draw((2.94,-4.24)--(5.720585931236688,0.9279576903792971), linewidth(1.) + qqzzcc); draw((8.133298946500023,0.9578157476001203)--(11.040923834545335,-4.139748604221387), linewidth(1.) + qqzzcc); draw((2.94,-4.24)--(8.133298946500023,0.9578157476001203), linewidth(1.) + ccccff); draw((5.720585931236688,0.9279576903792971)--(11.040923834545335,-4.139748604221387), linewidth(1.) + ccccff); draw(circle((7.009262499434532,-5.709075934721522), 4.3263242355702), linewidth(1.) + qqccqq); draw(circle((6.944189219675456,-0.4507582933744881), 1.8433835117349662), linewidth(1.) + ttffcc); draw((4.374406723856454,-0.2667693776957385)--(5.720585931236688,0.9279576903792971), linewidth(1.) + zzwwff); draw((4.374406723856454,-0.2667693776957385)--(8.380754882891011,-1.6058954569210449), linewidth(1.) + zzffqq); draw((2.94,-4.24)--(4.374406723856454,-0.2667693776957385), linewidth(1.) + zzffqq); draw((4.374406723856454,-0.2667693776957385)--(5.536649473250011,-1.6410921261999398), linewidth(1.) + zzwwff); draw((5.536649473250011,-1.6410921261999398)--(5.720585931236688,0.9279576903792971), linewidth(1.) + blue); draw((8.380754882891011,-1.6058954569210449)--(8.133298946500023,0.9578157476001203), linewidth(1.) + blue); draw((8.380754882891011,-1.6058954569210449)--(5.536649473250011,-1.6410921261999398), linewidth(1.) + blue); draw((5.536649473250011,-1.6410921261999398)--(11.040923834545335,-4.139748604221387), linewidth(1.) + ccwwff); draw((8.380754882891011,-1.6058954569210449)--(2.94,-4.24), linewidth(1.) + ccwwff); draw(circle((5.659297971765954,-0.35876383553511265), 1.2881803054027774), linewidth(1.) + ffqqff); draw((4.374406723856454,-0.2667693776957385)--(6.941519152374644,-0.23500058078616584), linewidth(1.) + afeeee); draw(circle((5.691834611645495,-2.987922656208629), 3.02329148523356), linewidth(1.) + ffdxqq); draw((4.374406723856454,-0.2667693776957385)--(5.536649473250011,-1.6410921261999398), linewidth(1.) + ffcqcb); draw((5.536649473250011,-1.6410921261999398)--(2.94,-4.24), linewidth(1.) + ffcqcb); draw((2.94,-4.24)--(4.374406723856454,-0.2667693776957385), linewidth(1.) + ffcqcb); draw((4.374406723856454,-0.2667693776957385)--(8.380754882891011,-1.6058954569210449), linewidth(1.) + ffcqcb); draw((8.380754882891011,-1.6058954569210449)--(5.720585931236688,0.9279576903792971), linewidth(1.) + ffcqcb); draw((5.720585931236688,0.9279576903792971)--(4.374406723856454,-0.2667693776957385), linewidth(1.) + ffcqcb); /* dots and labels */ dot((2.94,-4.24),dotstyle); label("$D$", (2.984486772486775,-4.140804232804229), NE * labelscalefactor); dot((11.040923834545335,-4.139748604221387),dotstyle); label("$C$", (11.087005291005303,-4.038370370370367), NE * labelscalefactor); dot((5.720585931236688,0.9279576903792971),dotstyle); 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/* end of picture */ [/asy]$

Sketch for the solution:

First of we redefine $P$ as the intersection of the tangents $PD$ , $PE$ .

1- Show that $PDEH$ cyclic.

2- Show that $PED$ congruent to $PAF$

Then a spiral similarity centered at $P$ sending $DE$ to $AF$ does the job.

This post has been edited 5 times. Last edited by Ywgh1, Jun 9, 2024, 3:08 PM

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MathLuis

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MathLuis

#79 h Jul 6, 2024, 12:23 AM • 3 Y

Y by OronSH, teomihai, kingu

Evil MathLuis be like:
Let $F$ midpoint of $BD$ , now take $(ECDF)$ as the unit circle and let $E=e, C=c, D=d, F=f$
$CD \parallel EF \implies \frac{d-c}{f-e} \in \mathbb R \implies \frac{d-c}{f-e}=\frac{\overline{d}-\overline{c}}{\overline{f}-\overline{e}}=\frac{\frac{c-d}{cd}}{\frac{e-f}{ef}} \implies f=\frac{cd}{e}$ Now redefine $P=p=\frac{2ed}{e+d}$ and due to reflection $A=a=2e-c$ , now we compute $\text{arg}(\angle FAP)$
$\frac{f-a}{p-a}=\frac{\frac{cd}{e}-(2e-c)}{\frac{2ed}{e+d}-(2e-c)}=\frac{e+d}{e} \cdot \frac{cd-2e^2+ec}{2cd-2e^2+ec-2cd+cd}=\frac{e+d}{e}$ $\angle FAP=\angle DCE \iff \frac{d-c}{e-c} : \frac{f-a}{p-a} \in \mathbb R \iff \frac{d-c}{e-c} \cdot \frac{e}{e+d}=\frac{d-c}{e-c} \cdot \frac{ec}{dc} \cdot \frac{d}{e+d}=\frac{d-c}{e-c} \cdot \frac{e}{e+d}$ Therefore $\angle FAP=\angle DCE=\angle FEA$ thus $AP$ is tangent to $(ABE)$ as desired thus we are done :diablo:

This post has been edited 1 time. Last edited by MathLuis, Jul 6, 2024, 12:24 AM

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SomeonesPenguin

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SomeonesPenguin

#80 h Sep 17, 2024, 10:58 AM • 1 Y

Y by teomihai

Let $AC\cap BD=\{O\}$ and let $a:=\angle ACD$ .

Claim: $\angle APD=\angle DEC+\angle AEB$

Proof: $\angle DEC=180^\circ-\angle EDC-a$ but notice that $\angle EDC=\angle PDO$ so $\angle DEC=180^\circ-\angle PDO-a$ we similarly get $\angle AEB=180^\circ-\angle PAO-a$ , hence their sum is equal to $360^\circ-\angle PDO-\angle PAO-2a$ and the conclusion follows from the quadrilateral $PDOA$ and noting that $\angle AOD=2a$ .

Claim: $\frac{\sin(\angle DPO)}{\sin(\angle APO)}=\frac{\sin(\angle DEC)}{\sin(\angle AEB)}$

Proof: Trig Ceva in $\triangle APD$ with point $O$ gives: $\frac{\sin(\angle DPO)}{\sin(\angle APO)}=\frac{\sin(\angle DAO)}{\sin(\angle PAO)}\cdot\frac{\sin(\angle PDO)}{\sin(\angle ADO)}$
We also have that $\angle PDO=\angle EDC$ and $\angle PAO=\angle ABE$ and $\frac{\sin(\angle DAO)}{\sin(\angle ADO)}=\frac{DO}{AO}=\frac{DC}{AB}$ so LOS in $\triangle DEC$ and $\triangle ABE$ gives the desired conclusion.

Finally, from our claims we can conclude that $\angle DPO=\angle DEC$ and $\angle APO=\angle AEB$ so $PDOE$ is cyclic. This gives $a=\angle DCA=\angle PDE=\angle POA$ so $\angle POD=\angle PED=a$ , therefore $PE$ is tangent to $\Omega$ . $\blacksquare$

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cj13609517288

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cj13609517288

#81 h Sep 25, 2024, 3:41 PM

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I bashed this on paper, so here's a summary:

Redefine $P$ as the pole of $DE$ wrt $(CDE)$ . Then we want to show that $\angle PAF=\angle ABF$ . Now we can just use complex with $(CDE)$ as the unit circle. Eventually we get that we want to prove
$\frac{de(d-f)}{(d+e)(2de-ef-2df+d^2)}\in\mathbb{R}.$ This is not hard to do.

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peppapig_

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peppapig_

#82 h Jan 1, 2025, 5:33 PM • 1 Y

Y by teomihai

Let $(CDE)$ be the unit circle so that $c=\frac{1}{d}$ .

Note

* $a=2e-c$ ,
* $b=\overline a=\frac{2c-e}{ec}$ .

Let $P'=DD\cap EE$ . Note
$p'=\frac{2de}{d+e}=\frac{2e}{1+ec}.$ Then,
$P=P'\iff P'\in AA \iff \angle P'AE=\angle ABE,$ so
$\frac{\left(\frac{p'-a}{e-a}\right)}{\left(\frac{a-b}{e-b}\right)}=\frac{c(1+ec-2e^2)(-2c+e+e^2c)}{(1+ec)(c-e)(2e^2c-ec^2-2c+e)}.$ Conjugating,
$\left(\overline{\frac{c(1+ec-2e^2)(-2c+e+e^2c)}{(1+ec)(c-e)(2e^2c-ec^2-2c+e)}}\right)=\frac{\left(\frac{1}{c}\right)\left(\frac{e^2c+e-2c}{e^2c}\right)\left(\frac{-2e^2+ec+1}{e^2c}\right)}{\left(\frac{1+ec}{ec}\right)\left(\frac{e-c}{ec}\right)\left(\frac{2c-e-2e^2c+ec^2}{e^2c^2}\right)},$ $=\frac{c(1+ec-2e^2)(-2c+e+e^2c)}{(1+ec)(c-e)(2e^2c-ec^2-2c+e)}.$ They are equal, so $P=P'$ .

This post has been edited 2 times. Last edited by peppapig_, Jan 1, 2025, 5:34 PM
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ihatemath123

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ihatemath123

#83 h Mar 17, 2025, 8:55 PM • 1 Y

Y by OronSH

This problem is a corollary of the main claim in the solution to 2019 TST #1. See the diagram of Evan Chen's solution for reference. In that diagram, we claim $X$ is in fact the spiral center from $A \cup (AMN)$ to $M \cup (MBY)$ . Indeed, this is easy to check by adding in the centers $O_1$ and $O_2$ of $(AMN)$ and $(MBY)$ and showing with lengths that $\triangle XAO_1 \sim \triangle XMO_2$ . In the referenced solution, it is shown with symmedians that the second tangent from $X$ to $(AMN)$ lies on $(BMY)$ as well.

Analogously, in the RMM problem, it follows that $P$ is the center of spiral similarity sending $A \cup (ABE)$ to $E \cup (CED)$ . Moreover, the second tangent from $P$ to $(ABE)$ is $F$ , the second intersection of $(ABE)$ and $(CED)$ . Clearly, line $EF$ is the midline of the trapezoid. But since $\angle ECD = \angle FBA$ , arcs $ECD$ and $ABF$ are similiar, implying $\triangle PAF \sim \triangle PED$ , which finishes.

This post has been edited 1 time. Last edited by ihatemath123, Mar 18, 2025, 4:07 PM

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