Let $ABC$ be an acute-angled triangle with $AC > BC$, with incircle $\tau$ centered at $I$ which touches $BC$ and $AC$ at points $D$ and $E$, respectively. The point $M$ on $\tau$ is such that $BM \parallel DE$ and $M$ and $B$ lie on the same halfplane with respect to the angle bisector of $\angle ACB$. Let $F$ and $H$ be the intersections of $\tau$ with $BM$ and $CM$ different from $M$, respectively. Let $J$ be a point on the line $AC$ such that $JM \parallel EH$. Let $K$ be the intersection of $JF$ and $\tau$ different from $F$. Prove that $ME \parallel KH$.
Problem
Source: European Mathematical Cup 2022, Junior Division, Problem 3
Tags: geometry, incircle, Parallel Lines, angle bisector
on_gale
20.12.2022 03:37
I initially thought in Pascal's theorem because I felt that I needed something to deal with a circle and collinear points or concurrent lines. But then: Let $\alpha = \angle{EMH}$. Since $EC$ is tangent to the incircle, then $\alpha = \angle{CEH}$. Let $G = MF\cap EC$. $ME \| KH \iff \angle{EMH} = \angle{KEM} = \angle{KFM} = \angle{JFG} \iff GJ^2 = GF \cdot GM$ So we want to prove that $G$ is the midpoint of $JE$ because we know that $GE^2 = GF \cdot GM$ Then I just wanted to play with the cross ratio: $(MJ, ME; MG, p(M, IE)) = -1 \iff (p(I, HE), p(I, ME); p(I, DE), IE) = -1 \iff (EH, EM; ED, EC) = -1$ And the latter is true since $CD, CE$ are tangents to the incircle and $M, H, C$ are collinear. We're done! (The first equality comes from lemma applied around point $I$ and the second one comes from lemma around point $E$)
Notation: $p(X, YZ)$ is the line passing through $X$ perpendicular to $YZ$
Lemma
Let $P, X, Y, Z, W$ be coplanar distinct points. Let $Q$ be any point in the same plane.
Then $(PX, PZ; PY, PW) = (p(Q, PX), p(Q, PZ); p(Q, PY), p(Q, PW))$
The proof consists in looking at the angles formed by those pencils (they're the same!)
Assassino9931
20.12.2022 03:45
(Synthetic blah, but no idea if it is really *that* hard.)
Our main focus (equivalent to the desired statement) is to show $KE = MH$ by finding convenient expressions for both segments.
First part is on $MH$. Let $BF \cap AC = T$ -- then $BCT$ is isosceles and $BM = FT$ by symmetry. Now Stewart's theorem (which for the isosceles $BCT$ is just Power of a Point with respect to the circle centered at $C$ with radius $CB = CT$) we obtain $CM^2 = BC^2 - BM \cdot MT = BC^2 - BM \cdot BF = BC^2 - BD^2 = a^2 - \frac{(a+c-b)^2}{4} = \frac{(a+b-c)(3a+c-b)}{4}$ where $BC = a$, $AC = b$, $AB = c$. (We have used $BD^2 = BM \cdot BF$ from Power of a Point and $BD = \frac{a+c-b}{2}$ from the incircle.) On the other hand, Power of a Point also gives $(CM - MH) \cdot CM = CH \cdot CM = CD^2 = \frac{(a+b-c)^2}{4}$, whence
$$ MH = \frac{CM^2 - \frac{(a+b-c)^2}{4}}{CM} = (a+c-b)\sqrt{\frac{a+b-c}{3a+c-b}}.$$(Also, $CH = CM - MH = \frac{(a+b-c)^{3/2}}{2\sqrt{3a+c-b}}$.)
Now we move on to $KE$. Observe that $\triangle KEJ \sim \triangle EFJ$ and so $KE = \frac{EF \cdot EJ}{FJ}$ so we no longer need $K$!
For $EJ$ we apply Thales' theorem on $EH \parallel JM$, to obtain $$EJ = CE \cdot \frac{MH}{CH} = \frac{a+b-c}{2} \cdot (a+c-b)\sqrt{\frac{a+b-c}{3a+c-b}} \cdot \frac{2\sqrt{3a+c-b}}{(a+b-c)^{3/2}} = a+c-b$$(wow, so $T$ seems to be the midpoint of $EJ$, this might be because of some stupid synthetic reason but I am blind enough to see it I guess).
Before we move on, let us compute $BM$. If $BM = x$ and $MF = y$, then $2x + y = BT = 2a\sin \frac{\gamma}{2}$ (as usual $\angle ACB = \gamma$) and $x(x+y) = BD^2 = \frac{(a+c-b)^2}{4}$, hence $x(2a\sin \frac{\gamma}{2} - x) = \frac{(a+c-b)^2}{4}$, i.e. $x\left(\sqrt{\frac{a(c-a+b)(a+c-b)}{b}} - x\right) = \frac{(a+c-b)^2}{4}$. (We used $\sin \frac{\gamma}{2} = \sqrt{\frac{1-\cos\gamma}{2}}$ and the Law of Cosines in $ABC$.)
For $EF = DM$ we focus on $\triangle BMD$. The Law of Cosines in gives
$$ EF = DM = \sqrt{BD^2 + BM^2 - 2 \cdot BM \cdot BD \cdot \cos\left(90^{\circ} - \frac{\gamma}{2}\right)} $$$$ = \sqrt{x^2 + \frac{(a+c-b)^2}{4} - (a+c-b)x\sin \frac{\gamma}{2}} $$$$ = \sqrt{2ax\sin \frac{\gamma}{2} - (a+c-b)\sin \frac{\gamma}{2}} = \sqrt{(a+b-c)x\sin \frac{\gamma}{2}}. $$For $FT$ focus on triangle $FTJ$. One way would be to note that $\angle CTF = 90^{\circ} - \frac{\gamma}{2}$ and use the Law of Cosines, but in my opinion it is better to use that $FT$ is a median (established above when computing $EJ$) and hence
$$ 4x^2 = 4FT^2 = 2EF^2 + 2FJ^2 - JE^2, \mbox{ i.e. } $$$$ FJ = \sqrt{2x^2 + \frac{(a+c-b)^2}{2} - (a+b-c)x\sin \frac{\gamma}{2}} = \sqrt{4ax \sin \frac{\gamma}{2} - (a+c-b)x\sin \frac{\gamma}{2}} $$$$ \sqrt{(3a+c-b)x\sin \frac{\gamma}{2}}. $$Therefore $KE = (a+c-b)\sqrt{\frac{a+b-c}{3a+c-b}}$, the end!
Here is a simpler solution (which completely avoids trigonometry) when you combine @on_gale's approach with mine. Reduce firstly to proving $G$ (in my notation $T$) to be the midpoint of $JE$ as they did and then take only the first two of my calculations, which are the one with Stewart/PoP and the one with Thales. Amazing!
VicKmath7
20.12.2022 10:31
Here is a simple synthetic solution using only angle chasing and similar triangles. We want $ME \parallel HK$, so we need $\angle MFK = \angle CEH$. This motivates to let $MF \cap EH=G$ and we want $GEFJ$ being cyclic. We have that $\angle EGF = \angle HED= \angle HMD =\angle EMN$ (where $N$ is midpoint of $ED$ and it lies on $FH$ by the harmonic quadrilateral $MEHD$). Thus, we want $\triangle EMN \sim \triangle EJF \iff \frac{EF}{EJ}=\frac{EN} {EM} \iff \frac {EJ} {EM}=\frac {DM} {DN}$. The last one follows from the similarity of the triangles $DNM$ and $EMJ$ ($\angle NDM =\angle JEM$ and $\angle EMJ = \angle MEH = \angle MND$), done.
Assassino9931
20.12.2022 10:45
@above well the need of harmonic quadrilateral is a bit out of the ordinary similarity stuff, especially for juniors Anyway, nice solution!
Kimchiks926
20.12.2022 14:16
A bit boring problem. Step 1: $\triangle JEF \sim \triangle MHD$ Proof: Observe that $EDFM$ is isosceles trapezium, therefore by symmetry we have $\triangle CEF = \triangle CDM$. Moreover $\angle HDC = \angle CMD = \angle CFE$, which implies that $\triangle CHD \sim \triangle CEF$. This means that we have: $$ \frac{CE}{EH}=\frac{EF}{HD} $$Since $EH \parallel JM$ we have $\frac{CE}{CH}=\frac{JE}{MH}$. We conclude that: $$ \frac{CE}{EH}=\frac{EF}{HD} =\frac{JE}{MH} $$Moreover, $\angle JEF = \angle EMF = \angle DFM = \angle DHM $. This implies that $\triangle JEF \sim \triangle MHD$ as desired. We define point $T \neq M$ to be the intersection of $JM$ and $\tau$. Step 2: $CJ \parallel KT$ Proof: Observe that since $EDMF$ and $EHMT$ are isosceles trapezium, then $\widehat{EF}=\widehat{DM}$ and $\widehat{ET}=\widehat{MH}$. Therefore $\widehat{FT}=\widehat{ET}-\widehat{EF}= \widehat{MH}-\widehat{DM}=\widehat{DH}$. Thus $\angle EJF = \angle DMH = \angle FKT$, which implies $CJ \parallel KT$ as desired. Step 3: Finish Finish Observe that: $$ \angle KTM = \angle CJM = \angle CEH = \angle HME $$This means that $\widehat{EH} = \widehat{KM}$, therefore $EHKM$ is isosceles trapezium, implying that $ME \parallel KH$ as desired.
Attachments:
MR_D33R
13.12.2023 02:36
Consider a homothety at $C$ sending $E$ to $J$. By $EH || JM$ we get that $H$ gets sent to $M$, let's say that $D$ gets sent to $D'$. We have that $MF,D'J$ are both perpendicular to $CI$, so by symmetry $JD'MF$ is a trapezium. Finally we have \[\sphericalangle{HDE} =\sphericalangle{MD'J} = \sphericalangle{D'JF}=\sphericalangle{MFK}\implies EH=MK \implies HK || EM.\]