Perpendicular

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Source: Iranian National Olympiad (3rd Round) 2006

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1275 posts

#1 h Sep 21, 2006, 5:22 PM • 2 Y

Y by Adventure10, Mango247

$ABC$ is a triangle and $R,Q,P$ are midpoints of $AB,AC,BC$ . Line $AP$ intersects $RQ$ in $E$ and circumcircle of $ABC$ in $F$ . $T,S$ are on $RP,PQ$ such that $ES\perp PQ,ET\perp RP$ . $F'$ is on circumcircle of $ABC$ that $FF'$ is diameter. The point of intersection of $AF'$ and $BC$ is $E'$ . $S',T'$ are on $AB,AC$ that $E'S'\perp AB,E'T'\perp AC$ . Prove that $TS$ and $T'S'$ are perpendicular.

This post has been edited 3 times. Last edited by Omid Hatami, Sep 23, 2006, 5:37 AM

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594 posts

#2 h Sep 21, 2006, 6:47 PM • 1 Y

Y by Adventure10

easy problem

. First of all I think that there's a mistake , $P,Q,R$ must be the midpoints of $BC,CA,AB$ respectively...
Now because: $PRAQ$ is a parallelogram and because $\angle ETP=\angle ESP=90$ so $ETPS$ is cyclic and $\angle TES=180-\angle TPS=180-\angle RAQ =180-\angle BAC = \angle T^{'}AS^{'}$ and because $AT^{'}E^{'}S^{'}$ is cyclic too, so the quadrilaterals $AT^{'}E^{'}S^{'}$ and $ETPS$ are similar. Now assume that $EP\cap TS=D$ and $ST\cap S^{'}T^{'}=G$ and $S^{'}T^{'}\cap AE^{'}=J$ .
because $AT^{'}E^{'}S^{'}$ and $ETPS$ are similar so we have $\angle ADG=\angle EDT=\angle T^{'}JA=180-\angle AJG$ , so $AJGD$ is cyclic too so because $FF^{'}$ is a diameter of the circumcircle of $ABC$ we have $\angle JAD=\angle F^{'}AF=90$ , and because $AJGD$ is cyclic we have : $\angle JGD=90$ and that means $SG$ or $ST$ is perpendicular to $S^{'}T^{'}$ .

This post has been edited 2 times. Last edited by sinajackson, Sep 23, 2006, 12:37 PM

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#3 h Sep 22, 2006, 7:58 AM • 2 Y

Y by Adventure10, Mango247

sinajackson wrote:

Easy problem : $PRAQ$ is a parallelogram and because $\angle ETP=\angle ESP=90$ so $ETPS$ is cyclic and
$\angle TES=\ldots =\angle T^{'}AS^{'}$ and because $AT^{'}E^{'}S^{'}$ is cyclic too, so the quadrilaterals $AT^{'}E^{'}S^{'}$ and $ETPS$ are similar.

And yet, why ... are similarly ?

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#4 h Sep 22, 2006, 11:03 AM • 1 Y

Y by Adventure10

Well because $\angle AT^{'}E^{'}=\angle ETP,\angle AS^{'}E^{'}=\angle ESP, \angle T^{'}AS^{'}=\angle TES$ and $\angle T^{'}E^{'}S^{'}=\angle TPS$ , .... is there a problem?

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#5 h Sep 22, 2006, 4:34 PM • 3 Y

Y by Adventure10, Mango247, Mango247

Sorry, I think was sleeping then. It is all O.K.

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1275 posts

#6 h Sep 23, 2006, 5:39 AM • 2 Y

Y by Adventure10, Mango247

To Sinajackson :
I corrected the problem, but it's different with what you said.

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#7 h Sep 23, 2006, 6:12 AM • 2 Y

Y by Adventure10, Mango247

And yet, I am right ! $A=A'\ ,\ B=B'\ ,\ C=C'\ ,\ D=D'\not\Longrightarrow ABCD\sim A'B'C'D'$ , i.e. exist at least two quadrilaterals with the same angles and which yet aren't similarly.

This post has been edited 1 time. Last edited by Virgil Nicula, Sep 23, 2006, 6:27 AM

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#8 h Sep 23, 2006, 6:22 AM • 2 Y

Y by Adventure10, Mango247

Yes, SinaJackson's solution is not right.

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#9 h Sep 23, 2006, 12:28 PM • 1 Y

Y by Adventure10

Hello Mr Hatami and Virgil Nicula.... I think my solution is correct, I agree that we can find two quadrileterals that have the same angles and are not similar.... but here there's a difference, we have two right angles in these quadrilaterals...
I think my solution is correct....

( sorry for having much confidence

)

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7054 posts

#10 h Sep 23, 2006, 5:29 PM • 2 Y

Y by Adventure10, Mango247

Example for Sinajackson.

Quote:

Let $ABCD$ be a fixed convex quadrilateral inscribed in the circle $w(O)$ with the diameter $[AC]$ , i.e. $B=D=90^{\circ}$
Let $\{X,Y\}\subset w$ be two mobile points so that $XY=BD$ , the line $AC$ does not separate the points $B$ , $X$ and the line $XY$ separates the points $A$ , $O$ .
Then the convex cyclic quadrilaterals $ABCD$ and $AXCY$ , where $B=X=D=Y=90^{\circ}$ , have the same values of the its (corresponding to the writing) angles, i.e. and $\widehat{BAD}\equiv\widehat{XAY}$ , $\widehat{BCD}\equiv\widehat{XCY}$ and yet they never aren't similarly, i.e. $ABCD\not\sim AXCY\ .$

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509 posts

#11 h Sep 23, 2006, 7:41 PM • 1 Y

Y by Adventure10

But they have all their edges perpendicular ...

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#12 h Sep 24, 2006, 4:12 AM • 1 Y

Y by Adventure10

sorry , but I still think my solution is ok, and I think it is completely obvious that we can use the similarity of the two cyclic quadrilaterals in my solution....
your example is a completely special one that will not occur every time......

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#13 h Sep 24, 2006, 4:19 AM • 1 Y

Y by Adventure10

Are agree you with the my example ? I didn't say that the your proof is wrong, but that you used a false assertion.

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#14 h Sep 24, 2006, 4:04 PM • 2 Y

Y by Adventure10, Mango247

Hm...what do you need similar quadrilaterals for?

My solution:

We have, since $TESP$ is cyclic $\angle T'AS'=180-\angle CAB=180-\angle TPS=\angle TES.$ Also, since $E'A\perp AP$ and $T'A\perp T'E'$ , we have $\angle EST=\angle EPT=\angle PAQ=\angle AE'T'=\angle AS'T'.$ Hence the triangles $AT'S'$ and $ETS$ are similar (with the same orientation).
So there exists a spiral similarity taking $ETS$ to $AT'S'$ . However the corresponding sides $ES$ and $AS'$ , as well as $ET$ and $AT'$ are perpendicular, so the third pair of sides, $TS$ and $T'S'$ must also be perpendicular.

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708 posts

#15 h Sep 24, 2006, 9:00 PM • 1 Y

Y by Adventure10

After All this is an Iranian Problem

:

sinajackson wrote:

Well because $\angle AT^{'}E^{'}=\angle ETP,\angle AS^{'}E^{'}=\angle ESP, \angle T^{'}AS^{'}=\angle TES$ and $\angle T^{'}E^{'}S^{'}=\angle TPS$ , .... is there a problem?

Yes there is.
Sina , Livi (Virgil Nicula) gave you a counterexample.

Here is another one :

Every rectangle has $4$ right angles , Do you think they are all Similar? I dont think so.

If you still dont believe me , Prove your claime rigorously.

Yimin ge ,I solved the problem like you.

I wonder how they created this problem ?

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718 posts

Gibby

#16 h Jun 6, 2014, 4:30 AM • 2 Y

Y by Adventure10, Mango247

Can somebody please post a diagram for this problem?

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290 posts

jred

#17 h Aug 28, 2016, 7:38 AM • 2 Y

Y by Adventure10, Mango247

I think the difficulty of this problem is that we have to consider various configurations. For a specific case, the problem is easy.

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