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Source: Iran Second Round 2015 - Problem 3 Day 1

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128 posts

#1 h May 7, 2015, 10:45 AM • 7 Y

Y by Dadgarnia, Sayan, AdithyaBhaskar, bgn, Adventure10, Mango247, Rounak_iitr

Consider a triangle $ABC$ . The points $D,E$ are on sides $AB,AC$ such that $BDEC$ is a cyclic quadrilateral. Let $P$ be the intersection of $BE$ and $CD$ . $H$ is a point on $AC$ such that $\angle PHA = 90^{\circ}$ . Let $M,N$ be the midpoints of $AP,BC$ . Prove that: $ACD \sim MNH$ .

This post has been edited 5 times. Last edited by AHZOLFAGHARI, Sep 11, 2015, 11:39 AM

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2312 posts

#3 h May 7, 2015, 11:21 AM • 11 Y

Y by Amir.S, AdithyaBhaskar, ILIILIIILIIIIL, M.Sharifi, kun1417, enhanced, aops29, hakN, seyyedmohammadamin_taheri, Adventure10, Mango247

AHZOLFAGHARI wrote:

Consider the triangle $ABC$ . The points $D,E$ are on sides $AB,AC$ such that $BDEC$ is cycle . The common point of $BE$ and $CD$ is $P$ . The point $H$ is on $AC$ such that $\angle PHA = 90$ . If $M,N$ in the midpoint of $AP,BC$ prove that : $AC\color{red}{ D }\normalcolor \sim MNH$ .

Typo corrected

My solution:

Let $F, G$ be the midpoint of $CP, CA$ , respectively .

Since $F, G, H, M$ are concyclic ( 9 point circle of $\triangle APC$ ) ,
so combine $\angle FMG=\angle ECD=\angle EBD=\angle FNG \Longrightarrow F, G, H, M, N$ are concyclic ,
hence from $\angle ACD=\angle HGM=\angle HNM , \angle MHN=\angle MFN=\angle AEB=\angle ADC \Longrightarrow \triangle ACD \sim \triangle MNH$ .

Q.E.D

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128 posts

#4 h May 7, 2015, 11:26 AM • 2 Y

Y by Adventure10, Mango247

TelvCohl wrote:

AHZOLFAGHARI wrote:

Consider the triangle $ABC$ . The points $D,E$ are on sides $AB,AC$ such that $BDEC$ is cycle . The common point of $BE$ and $CD$ is $P$ . The point $H$ is on $AC$ such that $\angle PHA = 90$ . If $M,N$ in the midpoint of $AP,BC$ prove that : $AC\color{red}{ D }\normalcolor \sim MNH$ .

Typo corrected

My solution:

Let $F, G$ be the midpoint of $CP, CA$ , respectively .

Since $F, G, H, M$ are concyclic ( 9 point circle of $\triangle APC$ ) ,
so combine $\angle FMG=\angle ECD=\angle EBD=\angle FNG \Longrightarrow F, G, H, M, N$ are concyclic ,
hence from $\angle ACD=\angle HGM=\angle HNM , \angle MHN=\angle MFN=\angle AEB=\angle ADC \Longrightarrow \triangle ACD \sim \triangle MNH$ .

Q.E.D

Yes , thanks , edited .

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4384 posts

#6 h May 7, 2015, 6:46 PM • 2 Y

Y by Adventure10, Mango247

Another (more complicated idea):
Let $K$ the projection of $P$ onto $AB$ , $L, Q$ the midpoints of $DE, KH$ .
Known properties: $M,L,N$ determine the Newton-Gauss line of $ADPE$ and $Q$ belongs to this line.
Also $KN=NH$ , easy to prove: take $O', O"$ the midpoints of $BP, PC$ and see congruent triangles, so $MN\bot KH$ .

Next, use the following lemma:

If, on the sides $KP, PH$ of a triangle $KPH$ are constructed the similar triangles $BPK, CPH$ and $X$ is the intersection of the perpendicular bisectors of $KH, BC$ respectively, then $m(\widehat{KXH})=2m(\widehat{KBP})$

Proof
Let $O',O"$ the circumcenters of the triangles $BPK, CPH$ and $DYH$ an isosceles triangle similar to $BPK, CPH$ , with $P$ and $Y$ on the same side of $KH$ .
Known property: $PO'YO"$ is a parallelogram (a spiral similarity kills the problem), and $\angle DYH=\angle PO"H$
Likewise, for $\triangle BPC$ with same similar triangles $BPK, CPH$ we have $\triangle BO'P\sim\triangle PO"C$ (isosceles) and construct the isosceles triangle $BZC$ similar to $BO'P$ , with P and Z on the same side of $BC$ . By same property, $PO'ZO"$ is a parallelogram, hence $Z\equiv Y\equiv X$ .

Now to problem:
$X\equiv N$ and $\angle DNH=2\angle DCA$ .
Also $M$ is the circumcenter of $AKPH$ , so $\angle DMH=2\angle BAC$ , so $MKNH$ is a kite with vertices angles $M, N$ being double of $\angle BAC$ and $\angle ACD$ , hence we are done.

Best regards,
sunken rock

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804 posts

Sardor

#7 h May 8, 2015, 3:42 AM • 2 Y

Y by Adventure10, Mango247

It's easy from nine-point circle and midline !
Nice problem from Iran !

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824 posts

andria

#8 h Jun 13, 2015, 12:42 PM • 2 Y

Y by Adventure10, Mango247

My first solution:
Let $R$ midpoint of $DE$ and $H'$ be the projection of $P$ on $AB$ points $M,N,R$ lie on newton gauss line of quadrilateral $ADPE$ according to this problem we deduce that $RN\perp HH'$ because $MH=MH'$ we get that $MN$ is perpendicular bisector of $HH'$ note that $M$ is center of cyclic quadrilateral $AHPH'$ so $\angle HMH'=2\angle A\longrightarrow \angle NMH=\angle A$ from IMO shotlist G4 2009 $MP$ is tangent to $\odot (\triangle RPN)$ so $\triangle MPR\sim \triangle MNP\longrightarrow \frac{MN}{MP}=\frac{PN}{PR}$ because $MP=MH\longrightarrow \frac{MN}{MH}=\frac{PN}{PR}$ (1) but $\triangle PED\sim \triangle PCB\longrightarrow \frac{PN}{PR}=\frac{PE}{PC}=\frac{AD}{AC}$ (2) from (1),(2): $\frac{AD}{AC}=\frac{MN}{MH}\longrightarrow \triangle MNH\sim \triangle DCA$
DONE

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824 posts

andria

#9 h Jun 13, 2015, 12:44 PM • 2 Y

Y by Adventure10, Mango247

My second solution:
$S,T$ are reflections of $P$ throw $H$ and $N$ clearly $\triangle ASC=\triangle APC$ because $HN|| ST,HM|| SA\longrightarrow \triangle MHN\sim \triangle AST$ from IMO shortlist G2 2012 quadrilateral $ASCT$ is cyclic $\longrightarrow \angle ATS=\angle ACS=\angle ACD$ (1) quadrilateral $PCTB$ is parallelgram so $EB|| CT\longrightarrow \angle ADC=\angle AEB=\angle ACT=\angle AST$ (2) from (1),(2) $\triangle ACS\sim \triangle ADC\sim \triangle MNH$
DONE

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tranquanghuy7198

253 posts

tranquanghuy7198

#10 h Jun 13, 2015, 2:44 PM • 2 Y

Y by ILIILIIILIIIIL, Adventure10

My solution:

Lemma.
Given $\triangle{ABC} \sim \triangle{DEF}$ (same direction)
$X, Y, Z\in{AD, BE, CF}: \frac{XA}{XD} = \frac{YB}{YE} = \frac{ZC}{ZF}$
We will have: $\triangle{ABC} \sim \triangle{DEF} \sim \triangle{XYZ}$
Proof. Vector rotating.

Back to our main problem.
$X, Y, Z, S$ are the midpoints of $DC, CA, AD, DE$
$\Rightarrow \overline{M, N, S}, \overline{Y, M, Z}, \overline{Z, S, X}, \overline{X, N, Y}$
$Q$ is the reflection of $P$ WRT $CE$
We have: $\triangle{PDB} \sim \triangle{PEC} \Rightarrow \triangle{PDB} \sim \triangle{QEC}$ (same direction)
But $H, S, N$ are the midpoints of $PQ, DE, BC$ $\Rightarrow \triangle{PDB} \sim \triangle{QEC} \sim \triangle{HSN}$ (lemma)
$\Rightarrow \triangle{HSN} \sim \triangle{PEC}$
$\Rightarrow \frac{NH}{NS} = \frac{CP}{CE}$ (1)
On the other hand: $\frac{NS}{NM} = \frac{XS}{XZ}.\frac{YZ}{YM} = \frac{CE}{CA}.\frac{CD}{CP}$ (2)
(1), (2) $\Rightarrow \frac{NH}{NS}.\frac{NS}{NM} = \frac{CP}{CE}.\left(\frac{CE}{CA}.\frac{CD}{CP}\right)$
$\Rightarrow \frac{NH}{NM} = \frac{CD}{CA}$
$\Rightarrow \triangle{HNM} \sim \triangle{DCA}$ (s.a.s)
Q.E.D

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1198 posts

#11 h Mar 24, 2016, 7:53 PM • 2 Y

Y by Adventure10, Mango247

TelvCohl wrote:

AHZOLFAGHARI wrote:

Consider the triangle $ABC$ . The points $D,E$ are on sides $AB,AC$ such that $BDEC$ is cycle . The common point of $BE$ and $CD$ is $P$ . The point $H$ is on $AC$ such that $\angle PHA = 90$ . If $M,N$ in the midpoint of $AP,BC$ prove that : $AC\color{red}{ D }\normalcolor \sim MNH$ .

Typo corrected

My solution:

Let $F, G$ be the midpoint of $CP, CA$ , respectively .

Since $F, G, H, M$ are concyclic ( 9 point circle of $\triangle APC$ ) ,
so combine $\angle FMG=\angle ECD=\angle EBD=\angle FNG \Longrightarrow F, G, H, M, N$ are concyclic ,
hence from $\angle ACD=\angle HGM=\angle HNM , \angle MHN=\angle MFN=\angle AEB=\angle ADC \Longrightarrow \triangle ACD \sim \triangle MNH$ .

Q.E.D

How did you get $\angle HGM=\angle ACD$ and $\angle MFN=\angle AEB$ ? And also, how did you come up with this solution?

This post has been edited 2 times. Last edited by viperstrike, Apr 29, 2016, 12:01 PM

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2016 posts

PROF65

#13 h Mar 25, 2016, 1:19 AM • 2 Y

Y by Adventure10, Mango247

Let $K,L$ be the feet of $P$ on $AB,BC \ . \ MH=MK= \frac{AP}{2},\widehat{KMH}=2 \ \hat A ,HKNL$ are on the circle of the feet of $P$ and its isogonal thus $\widehat{HKN}=\widehat{HLN}=\widehat{HPC}=\widehat{BPK}=\widehat{BLK}=\widehat{NHK}$ hence $NH=NK$ so $MHN$ is congruent to $MKN$ then it suffices to prove that $\widehat{KNH}=2 \cdot \widehat{DCA}$ but $\widehat{KNH}=\widehat{KNP}+\widehat{PNH}=\widehat{KBP}+\widehat{PCH}=2 \cdot \widehat{DCA}$ .
R HAS

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1498 posts

suli

#14 h Mar 25, 2016, 2:08 AM • 1 Y

Y by Adventure10

1. Let $E'$ be reflection of $E$ over $H$ .

2. Triangles $ABE$ and $PCE'$ are similar by AA Similarity. They have same orientation.

3. Triangle $ABE$ and $MNH$ are similar by Mean Geometry / Spiral similarity theory,

4. $ACD ~ ABE$ by AA similarity and thus $ACD ~ MNH$ by transitive property.

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763 posts

#16 h Apr 11, 2017, 9:33 PM • 2 Y

Y by Adventure10, Mango247

TelvCohl wrote:

AHZOLFAGHARI wrote:

Consider the triangle $ABC$ . The points $D,E$ are on sides $AB,AC$ such that $BDEC$ is cycle . The common point of $BE$ and $CD$ is $P$ . The point $H$ is on $AC$ such that $\angle PHA = 90$ . If $M,N$ in the midpoint of $AP,BC$ prove that : $AC\color{red}{ D }\normalcolor \sim MNH$ .

Typo corrected

My solution:

Let $F, G$ be the midpoint of $CP, CA$ , respectively .

Since $F, G, H, M$ are concyclic ( 9 point circle of $\triangle APC$ ) ,
so combine $\angle FMG=\angle ECD=\angle EBD=\angle FNG \Longrightarrow F, G, H, M, N$ are concyclic ,
hence from $\angle ACD=\angle HGM=\angle HNM , \angle MHN=\angle MFN=\angle AEB=\angle ADC \Longrightarrow \triangle ACD \sim \triangle MNH$ .

Q.E.D

I think $\angle ACD=\angle HGM=\angle HNM$ not right... in part $\angle HGM$ , which has contradiction with the case when $G$ is closer than $H$ to $A$ and would rather be $\angle ACD=\pi - \angle HGM(=\angle AGM)=\angle HNM$ in this case

This post has been edited 4 times. Last edited by Skravin, Apr 11, 2017, 9:49 PM

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4 posts

spy.

#17 h Jun 9, 2017, 7:38 PM • 2 Y

Y by Adventure10, Mango247

great problem to work on

This post has been edited 5 times. Last edited by spy., Apr 26, 2022, 2:03 PM
Reason: nothing

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32 posts

#19 h Jun 10, 2017, 5:56 PM • 2 Y

Y by Adventure10, Mango247

Let $R$ be the reflection of $P$ over $N$ and $S$ be the reflection of $P$ over $H$ . As $PH \perp AC$ , reflection over $H$ is same as reflection over $AC$ .

Lemma 1: $AR$ and $AP$ are isogonal wrt $\angle BAC$ .
Proof: In $\triangle ADP$ and $\triangle ACR$ , $\angle ADP=\angle ADE+\angle PDE=\angle ADE+\angle PBC=\angle ACB+\angle BCR=\angle ACR$ . [by reflection we had that $PBRC$ as a parallelogram.]
And, $\frac{DP}{CR}=\frac{DP}{PB}=\frac{DE}{BC}=\frac{AD}{AC}$ . So, $\triangle ADP \sim \triangle ACR$ . or, $\angle DAP= \angle CAR \Rightarrow \angle DAR +\angle RAP = \angle CAP+\angle PAR \Rightarrow \angle DAR=\angle CAP$ .
Which implies $AR$ and $AP$ are isogonal wrt $\angle BAC$ .

Lemma 2: $A, R, C, S$ are concyclic.
Proof: $180^o-\angle ASC =\angle SAC+\angle SCA=\angle PAC+\angle PCA=\angle APD$ . And we've shown that $\triangle ADP \sim \triangle ACR$ , which implies $\angle APD=\angle ARC$ . So, $\angle ARC= 180^o-\angle ASC \Rightarrow \angle ARC+\angle ASC=180^o$ .
So, $A, R, C, S$ are concyclic.

Lemma 3: $\triangle ADC \sim \triangle ASR$ .
Proof: We've shown that $\angle ADC=\angle ACR$ . and from lemma 2 $\angle ACR=\angle ASR$ . So, $\angle ADC=\angle ASR$ . And from lemma 1, $\angle DAR=\angle PAC=\angle CAS=a$ [last one is from reflection]. So, $\angle DAC=\angle DAR+\angle RAM+\angle PAC= 2a+\angle RAP$ . And, $\angle SAR= \angle RAP+\angle PAC+ \angle CAS=\angle RAP+2a$ . So, $\angle DAC= \angle SAR$ .
Which implies $\triangle ADC \sim \triangle ASR$ .

Lemma 4: $\triangle ASR \sim \triangle MHN$ .
Proof: $N,H,M$ are the midpoint of the side $\overline{PR}, \overline{PS}, \overline{PA}$ . So, $MN \parallel AR, MH \parallel AS, HN \parallel SR$ .
So, $\triangle SAR \sim \triangle HMN$ .[Note this cam be done with a homothety center at $P$ with ratio $-\frac{1}{2}$ ].

From lemma 3, $\triangle ADC \sim \triangle ASR$ . From lemma 4 $\triangle ASR \sim \triangle MHN$ . So, $\triangle ADC \sim \triangle MNH$ .

$\mathbb Q. \exists. \mathbb D.$

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#20 h Jan 12, 2018, 5:33 PM • 2 Y

Y by Adventure10, Mango247

Sol

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Mahdi_Mashayekhi

694 posts

Mahdi_Mashayekhi

#21 h Jan 5, 2022, 7:34 AM

Y by

Nice one
Let S,Q be midpoints of CP and CA.

lemma1 : MHQSN is cyclic.
proof: we will prove MHQS and HQSN are cyclic.
∠AHP = 90 ---> ∠AHM = ∠MAH = 180 - ∠APC - ∠ACP = 180 - ∠QSC - ∠MSP = ∠QSM ---> MHQS is cyclic.
∠PHC = 90 ---> ∠SHQ = ∠SCH = ∠PBD = ∠SNQ ---> HQSN is cyclic.

∠ACD = ∠AQM = ∠HNM and ∠DAC = ∠NQC = ∠NMH so NHM and CDA are similar.

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AgentC

#22 h Apr 2, 2024, 7:18 AM

Y by

suli wrote:

1. Let $E'$ be reflection of $E$ over $H$ .

2. Triangles $ABE$ and $PCE'$ are similar by AA Similarity. They have same orientation.

3. Triangle $ABE$ and $MNH$ are similar by Mean Geometry / Spiral similarity theory,

4. $ACD ~ ABE$ by AA similarity and thus $ACD ~ MNH$ by transitive property.

Would someone please explain step 2?

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#23 h Apr 13, 2025, 12:49 PM

Y by

Let $F$ be the reflection of $E$ over $H$ .
Claim: $ADPF$ cyclic.
Proof: $\angle BEC = \angle BDC = \angle PFA$
Using $Spiral$ $homogeneity$ it's enough to prove that: $ACD\sim FPC$ .
Which is true because $ADPF$ is cyclic. $\blacksquare$

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