BrMO 1 2015/16 question 5

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AndrewTom

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AndrewTom

#1 h Nov 28, 2015, 9:19 AM • 1 Y

Y by Adventure10

Let $ABC$ be a triangle, and let $D, E$ and $F$ be the feet of the perpendiculars from $A, B$ and $C$ to $BC, CA$ and $AB$ respectively. Let $P, Q, R$ and $S$ be the feet of the perpendiculars from $D$ to $BA, BE, CF$ and $CA$ respectively. Prove that $P, Q, R$ and $S$ are collinear.

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eshan

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eshan

#3 h Nov 28, 2015, 11:04 AM • 5 Y

Y by Viswanath, AndrewTom, Euler149, aops1221math, Adventure10

There's the same problem in RMO 2006.

Proof.
Clearly, $AFDC,AEDB$ are cyclic.
Now, $DP\perp AB,DQ\perp BE,DS\perp AE,$ and $D\in \odot ABE.$
So by pedal line theorem, $P,Q,S$ are collinear.
Similarly, $DP\perp AF,DR\perp CF,DS\perp AC,$ and $D\in \odot ACF.$
So $P,R,S$ are collinear.
As $P,Q,S$ and $P,R,S$ are collinear, we can say that $P,Q,R,S$ are collinear. $\Square$

My 300th post!!

PS -
Actually, this problem can be solved using many ways,
one as Jean-Louis said. Using miquel's theorem.
Another method is my method.
But in my method, I can consider some other pair of cyclic quads, one of them is pointed out by Wolowizard.

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jayme

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jayme

#10 h Nov 28, 2015, 12:43 PM • 3 Y

Y by AndrewTom, Adventure10, Mango247

Dear Mathlinkers,
the points in question of the initial problem lay on the Miquel's line of...

Sincerely
Jean-Louis

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PROF65

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PROF65

#11 h Nov 28, 2015, 8:21 PM • 3 Y

Y by AndrewTom, Adventure10, Mango247

THE SIMSON LINE IN TWO CIRCLES

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Wolowizard

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Wolowizard

#12 h Nov 28, 2015, 8:49 PM • 3 Y

Y by AndrewTom, Adventure10, Mango247

Let $H$ be orthocentre of triangle $\triangle ABC$ . Note that $BDFH$ and $DHEC$ are cyclic so by Simson's theorem on $\triangle BFH$ and $\triangle HEC$ : $P,Q,R$ and $Q,R,S$ are collinear.

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EulerMacaroni

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EulerMacaroni

#13 h Nov 28, 2015, 9:49 PM • 3 Y

Y by AndrewTom, Adventure10, Mango247

Invert about $D$ with radius $1$ , and denote inverses with a $'$ .

Note that $\angle DQ'B'=\angle DBQ=90^{\circ}-\angle C$ and $\angle DQ'R'=\angle DRQ=180^{\circ}-\angle DRS=\angle C$ , so that $P'Q'\perp Q'R'$ . Moreover, from $P', A', S'$ collinear, we get that $Q'R' \parallel P'S'$ , hence $P'Q'R'S'$ is a rectangle. From $\angle DQ'P'=\angle DPQ=\angle DPR=\angle DR'P'$ , we get the desired cyclicity, so inverting back yields $P,Q,R,S$ collinear $.\:\blacksquare\:$

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Stefan4024

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Stefan4024

#14 h Nov 29, 2015, 12:55 PM • 3 Y

Y by AndrewTom, Adventure10, Mango247

Here's another solution. It's slight overkill, but also it reveals some additional properties of the problem.

It's fairly easy using inversion to prove that $\odot (DRSC)$ and $\odot (BFEC)$ are tangent at point $C$ . Similarly $\odot (BPQD)$ and $\odot (BFEC)$ are tangent at point $B$ . Now using homothety wrt points $B$ and $C$ we find that $PQ \parallel RS \parallel EF$ It's obvious that $\odot (DRSC)$ and $\odot (BPQD)$ are tangent at $D$ and have common tangent line in $AD$ . Now using this we have: $AP \cdot AB = AD^2 = AS \cdot AC$ . Hence $BPSC$ are concyclic. Now again from the power of point $A$ and Thales' Theorem we have that $PS \parallel EF$ . Now from $PS \parallel PQ \parallel RS$ , $P,Q,R,S$ are collinear.

The problem can be stated slightly different. Let $B,D,C$ be collinear points, such that $D$ is between $B$ and $C$ . Let $k_1, k_2, k_3$ be circles with diametar $BC, BD, DC$ respectively. Let $E,F$ be points on $k_1$ , both on the same side of $BC$ . Let $FB \cap k_2 \equiv P$ , $EB \cap k_2 \equiv Q$ , $EC \cap k_3 \equiv S$ , $FC \cap k_3 \equiv R$ . Then prove that $P,Q,R,S$ are collinear $\iff B,P,S,C$ are concyclic $\iff AD \perp BC$ , where $A \equiv BF \cap CE$

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AndrewTom

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AndrewTom

#15 h Nov 29, 2015, 8:44 PM • 1 Y

Y by Adventure10

Thanks. Are there any other ways of solving this problem?

How do you prove, using inversion, that $\odot (DRSC)$ and $\odot (BFEC)$ are tangent at point $C$ ?

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Reason: Added something.

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Stefan4024

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Stefan4024

#16 h Nov 30, 2015, 8:32 AM • 3 Y

Y by AndrewTom, Adventure10, Mango247

AndrewTom wrote:

Thanks. Are there any other ways of solving this problem?

How do you prove, using inversion, that $\odot (DRSC)$ and $\odot (BFEC)$ are tangent at point $C$ ?

Consider a circle centered at $C$ with radius $\sqrt{CD\cdot CB}$ . Then $B$ and $D$ are each other's inverses, whiles also $E$ and $A$ are inverse pair. Now let $S'$ be the inverse of $S$ . Since $D$ lies on the polar of $S'$ , going through $S$ , $S'$ must lines on the polar of $D$ , which is the perpendicular to $BC$ at $B$ . So by inverting through the mentioned circle. $\odot (DRSC)$ goes to $BS'$ , while $\odot (BFEC)$ goes to $AD$ . But they are parallel and hence they have only one intesecting point (at infinity), hence they are tangent at $C$ . Otherwise the lines would have to intersect at another point.

More easily we can prove that they are tangent at $C$ by proving that they have a common tangent line at $C$ . Draw the tangent line to $\odot (DRSC)$ at C. Then the angle between the chord $SC$ and the tangent line is $\angle SDC = \angle EBC$ . Hence by the Tangent and Chord Theorem they share a common tangent line, hence they are tangent to each other. There are lots of other way to prove this. It's easy to prove that $C$ is the center of homothety of the two circles, hence they have only one touching point.

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aayush-srivastava

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aayush-srivastava

#17 h Nov 30, 2015, 8:35 AM • 4 Y

Y by AndrewTom, bond98, Adventure10, Mango247

AndrewTom wrote:

Thanks. Are there any other ways of solving this problem?

How do you prove, using inversion, that $\odot (DRSC)$ and $\odot (BFEC)$ are tangent at point $C$ ?

Yup!
Just consider the circumcircles of the triangles $ACD$ and $BEC$ ...
Both pass through $F$ .So by the $pedal$ $line$ $theorem$ we have $P,Q,R$ and $Q,R,S$ are collinear.(My 100th post!!!!!!!

)