Fixed points $B$ and $C$ are on a fixed circle $\omega$ and point $A$ varies on this circle. We call the midpoint of arc $BC$ (not containing $A$) $D$ and the orthocenter of the triangle $ABC$, $H$. Line $DH$ intersects circle $\omega$ again in $K$. Tangent in $A$ to circumcircle of triangle $AKH$ intersects line $DH$ and circle $\omega$ again in $L$ and $M$ respectively. Prove that the value of $\frac{AL}{AM}$ is constant. Proposed by Mehdi E'tesami Fard
Problem
Source: Iran 3rd round 2012-Geometry exam-P1
Tags: ratio, geometry, circumcircle, trigonometry, parallelogram, geometry proposed
pi37
21.09.2012 01:53
Note that
\[
\angle AKH+\angle DAH=\angle AKD+\angle DAH
\]
where $DAH$ is directed, and it is easy to angle chase and compute that this is $90$. Then the center of the circle through $A,K,H$ lies on $AD$. This implies that $\angle DAM=90$, so $M$ is the midpoint of arc $BAC$. Now since $MD\parallel AH$,
\[
\frac{AL}{AM}=\frac{1}{\frac{AM}{AL}}=\frac{1}{1-\frac{LM}{AL}}=\frac{1}{1-\frac{MD}{AH}}
\]
$MD$ is fixed, so it suffices to show that $AH$ is fixed, but this follows from the identity $AH=2R\cos A$.
ACCCGS8
25.09.2012 16:57
I find the strange rule concerning analytic solutions in the Geometry Exam, rather surprising. I have found a trigonometric solution to this problem, and will type it up when I find the time. I am very interested to see if users think it is "bashy" enough to be forbidden. How analytic does a solution have to be to be forbidden? It is a very fine line to draw, and like goodar2006, I don't really agree with such rules. Any method should be allowed as long as it is used properly. EDIT: Posting the solution I found. Let $\angle BAC = \alpha, \angle CBA = \beta, \angle ACB = \gamma$ and $\angle DHX = x$. $\angle MAX = \angle HKA = \angle DXA \implies DX = MA$. $\angle HKA = \angle DCA = \angle DCB + \angle BCA = \angle DAB + \angle BCA = \frac{\alpha}{2} + \gamma \implies \angle HLA = \angle HKA - \angle KAL = \angle HKA - \angle AHK = \frac{\alpha}{2} + \gamma - x$. From $\triangle HAL$, $\frac{AL}{\sin x} = \frac{AH}{\sin (\frac{\alpha}{2} + \gamma - x)} = \frac{\cos \alpha}{\sin (\frac{\alpha}{2} + \gamma - x)} \implies AL = \frac{\cos \alpha \sin x}{\sin (\frac{\alpha}{2} + \gamma - x)}$. $AH = 2R \cos \alpha$ and we have let $R = \frac{1}{2}$. $\angle DAX = \angle DAC - \angle XAC = \frac{\alpha}{2} - (90 - \gamma) \implies DX = \sin (\frac{\alpha}{2} + \gamma - 90) = - \cos (\frac{\alpha}{2} + \gamma)$ $\frac{AL}{AM} = \frac{AL}{DX} = - \frac{\cos \alpha \sin x}{\sin (\frac{\alpha}{2} + \gamma - x) \cos (\frac{\alpha}{2} + \gamma)} = - \frac{\cos \alpha}{(\sin (\frac{\alpha}{2} + \gamma) \cot x - \cos (\frac{\alpha}{2} + \gamma))\cos (\frac{\alpha}{2} + \gamma)}$ Now consider $\triangle DAX$. Let $Q$ be the foot of the altitude from $D$ to $AX$. $DQ = - \cos (\frac{\alpha}{2} + \gamma) \sin (\frac{\alpha}{2} + \gamma)$. $QR = \cos (\beta - \gamma) - \cos \alpha + (\cos (\frac{\alpha}{2} + \gamma))^2$ Thus $\frac{AL}{AM} = - \frac{\cos \alpha}{( \sin (\frac{\alpha}{2}+\gamma) \frac{QR}{DQ} - \cos (\frac{\alpha}{2} + \gamma))\cos (\frac{\alpha}{2} + \gamma)} = \frac{\cos \alpha}{\cos (\beta - \gamma) - \cos \alpha + 2(\cos (\frac{\alpha}{2} + \gamma))^2}$ $\cos (\beta - \gamma) = -\cos (2(\frac{\alpha}{2} + \gamma)) = 1 - 2(\cos (\frac{\alpha}{2} + \gamma))^2$ so $\frac{AL}{AM} = \frac{\cos \alpha}{1 - \cos \alpha}$ which is a constant. goodar2006, what do you think of this solution? What score was deemed to be 'good' for Iran 3rd Round 2012 Geometry Exam?
vslmat
25.09.2012 19:13
Let $P$ be the midpoint of $BC$, $OD$ cut the cirle $\omega$ again at $M'$. Since $LA$ is tangent to the circumcircle of $AHK$, $\angle LAK = \angle KHA = \angle KDM'$, thus $L, A, M'$ collinear, $M'\equiv M$. That $LO\parallel HP$, it is a well known problem. In short, let $LO$ cut $AH$ at $N$, it is a well known fact that $OP = 1/2 AH = HN$, thus $NHPO$ is a parallelogram and $LO\parallel HP$. $\frac{AL}{AM} = \frac{HL}{HD} = \frac{PO}{PD}$ = const.
Attachments:
phuongtheong
29.09.2012 17:35
We have: $LA^2=LK.LH; LA.LM=LK.LD$ thus $\dfrac{LA^2}{LA.LM}=\dfrac{LK.LH}{LK.LD} \Rightarrow \dfrac{AL}{LM}=\dfrac{LH}{LD}$. That follows $DM \| AH$. So we obtain $\dfrac{LA}{LM}=\dfrac{AH}{DM}=\dfrac{AH}{2R}$. Because $B$ and $C$ are the fixed point so $AH=const$. QED.
sayantanchakraborty
31.08.2014 21:32
It took me 5 minutes to come up with this solution,so its an easy one i guess: Since $AL$ is tangent to the circumcirle of $AKH$ we have $\angle{AHK}=\angle{LAK}=\angle{LDM}=\angle{KDM} \implies DM \parallel AH \implies DM \perp BC$.Now since $D$ is the midpoint of arc BC we see that it must me the diameter of the circumcircle.Now the rest is easy: $\frac{AL}{AM}=\frac{AH}{MD}=\frac{2RcosA}{2R}=cosA$ which is independent of the position of $A$.
jred
11.10.2016 18:36
sayantanchakraborty wrote: It took me 5 minutes to come up with this solution,so its an easy one i guess: Since $AL$ is tangent to the circumcirle of $AKH$ we have $\angle{AHK}=\angle{LAK}=\angle{LDM}=\angle{KDM} \implies DM \parallel AH \implies DM \perp BC$.Now since $D$ is the midpoint of arc BC we see that it must me the diameter of the circumcircle.Now the rest is easy: $\frac{AL}{AM}=\frac{AH}{MD}=\frac{2RcosA}{2R}=cosA$ which is independent of the position of $A$. There's a minor error, it should be $\frac{AL}{LM}=\frac{AH}{MD}$, therefore $\frac{AL}{AM}\ne cosA$. Hmmm, maybe it's easy to solve, but also easy to make mistake.