In acute triangle $ABC$, $AB>AC$. Let $I$ the incenter, $\Omega$ the circumcircle of triangle $ABC$, and $D$ the foot of perpendicular from $A$ to $BC$. $AI$ intersects $\Omega$ at point $M(\neq A)$, and the line which passes $M$ and perpendicular to $AM$ intersects $AD$ at point $E$. Now let $F$ the foot of perpendicular from $I$ to $AD$. Prove that $ID\cdot AM=IE\cdot AF$.
Problem
Source: Korea National Olympiad 2019 P6
Tags: geometry, Korea, geometry solved, Angle Chasing, similar triangles, incenter, arc midpoint
IMO2021
16.11.2019 18:56
It's pretty easy. Let $K$ be a touchpoint of the incircle and $BC$. It is well-known that $M,K,F$ are collinear (even it is still very easy to calculate). It means $DKIF$ is a rectangle, and so $KF=ID$. Clearly $IFEM$ is cyclic and $\Delta AIE$ and $\Delta AFM$ are similar, so $AF*IE=AI*MF$. It's finished after noticing $AI*MF=KF*AM=ID*AM$ which is true by Thales on $\Delta MAF$ and points $I,K$.
sharp0407
27.11.2019 10:03
(1) LET (AI,BC)=X, (MF,BC)=Y C(A, X ; I, I_A)=-1 --> DX IS BISECT <IDI_A(WELL-KNOWN) (2) LET AI=a, XI=b, MX=C --> c=b^2/(a-b) --> (AM/I_AM)=(a/b)=(AF/DF) --> MF || DI_A --> <IFY=<FYD=<IDY=<I_ADY (3) The rest is easy
Devastator
27.11.2019 13:54
I think I have another solution
We first prove the following 2 claims
Claim 1: $MI^2 = MK \cdot MA$
ProofBy the Incenter-Excenter Lemma, $MI=MB=MC$
Note that
$$\angle MBK = \angle MBC = \angle MAC = \angle MAB$$Thus, $\triangle EDI$ and $\triangle EIA$ are similar
Therefore,
$$\frac{MB}{MK}=\frac{MA}{MB}$$$$MB^2 = MK \cdot MA$$$$MI^2 = MK \cdot MA$$Thus, the Claim is proven
Claim 2: $\triangle EDI$ and $\triangle EIA$ are similar
ProofIt suffices to show that
$$\frac{EI}{EA} = \frac{ED}{EA}$$or
$$ EI^2 = ED \cdot EA$$Because $\angle EDK + \angle EMK = 90 + 90 = 180$, then quadrilateral $DEMK$ is cyclic. Thus, $AD \cdot AE = AK \cdot AM$
ThErEfOrE,
$$ED \cdot EA$$$$=(EA-DA) \cdot EA$$$$=(EM^2+AM^2)-AK\cdot AM$$$$=EM^2 + AM \cdot (AM-AK)$$$$=EM^2 + AM\cdot KM$$$$=EM^2 + IM^2$$(by Claim 1)
$$=EI^2$$Thus, the Claim is proven
Since $\angle EFI + \angle EMI = 90+90=180$, then quadrilateral $FEMI$ is cyclic, so
$$AF \cdot ARE = AI\cdot AM$$$$\frac{AI}{AE} = \frac{AF}{AM}$$$$\frac{ID}{IE}=\frac{AF}{AM}$$$$ID \cdot AM=IE\cdot AF$$$Q.E.D.$
jayme
09.12.2019 12:32
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/78.1%20Relations%20metriques.pdf p. 3... Sincerely Jean-Louis
mathverse06
08.07.2022 03:10
IMO2021 wrote: It's pretty easy. Let $K$ be a touchpoint of the incircle and $BC$. It is well-known that $M,K,F$ are collinear (even it is still very easy to calculate). It means $DKIF$ is a rectangle, and so $KF=ID$. Clearly $IFEM$ is cyclic and $\Delta AIE$ and $\Delta AFM$ are similar, so $AF*IE=AI*MF$. It's finished after noticing $AI*MF=KF*AM=ID*AM$ which is true by Thales on $\Delta MAF$ and points $I,K$. What is the easiest way to prove M, K, F are colinear?
hef4875
08.07.2022 11:04
mathverse06 wrote: IMO2021 wrote: It's pretty easy. Let $K$ be a touchpoint of the incircle and $BC$. It is well-known that $M,K,F$ are collinear (even it is still very easy to calculate). It means $DKIF$ is a rectangle, and so $KF=ID$. Clearly $IFEM$ is cyclic and $\Delta AIE$ and $\Delta AFM$ are similar, so $AF*IE=AI*MF$. It's finished after noticing $AI*MF=KF*AM=ID*AM$ which is true by Thales on $\Delta MAF$ and points $I,K$. What is the easiest way to prove M, K, F are colinear? Prove $MG.MA=MB^2$ by 2 similar triangles and $MB=MI$ you will get $MI^2=MG.MA$. Caculating that you will get $IA/IG=MA/MI$, since $AF/FD=IA/IG$ by combining we have $MA/MI=AF/IH$ so $M,H,F$ are collinear
ike.chen
09.07.2022 15:58
Let the incircle touch of $ABC$ touch $BC$ at $K$. By this lemma, we know $M, K, F$ are collinear. It's clear that $EFIM$ is cyclic with diameter $EI$, so $$\angle AMF = \angle IMF = \angle IEF = \angle IED.$$Now, since $DFIK$ is a rectangle, we have $$\angle AFM = 180^{\circ} - \angle DFM = 180^{\circ} - \angle DFK = 180^{\circ} - \angle FDI = \angle IDE$$implying $AFM \sim IDE$, which suffices. $\blacksquare$ Remark: The lemma I cited, which I stumbled across during September of 2021, basically incinerates this problem.
mathverse06
13.07.2022 07:44
ike.chen wrote: Let the incircle touch of $ABC$ touch $BC$ at $K$. By this lemma, we know $M, K, F$ are collinear. It's clear that $EFIM$ is cyclic with diameter $EI$, so $$\angle AMF = \angle IMF = \angle IEF = \angle IED.$$Now, since $DFIK$ is a rectangle, we have $$\angle AFM = 180^{\circ} - \angle DFM = 180^{\circ} - \angle DFK = 180^{\circ} - \angle FDI = \angle IDE$$implying $AFM \sim IDE$, which suffices. $\blacksquare$ Remark: The lemma I cited, which I stumbled across during September of 2021, basically incinerates this problem. Thx