Point of concurrence on the altitude

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Point of concurrence on the altitude

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Source: Iberoamerican 2015 #4

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681 posts

fprosk

#1 h Nov 11, 2015, 7:06 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute triangle and let $D$ be the foot of the perpendicular from $A$ to side $BC$ . Let $P$ be a point on segment $AD$ . Lines $BP$ and $CP$ intersect sides $AC$ and $AB$ at $E$ and $F$ , respectively. Let $J$ and $K$ be the feet of the peroendiculars from $E$ and $F$ to $AD$ , respectively. Show that

$\frac{FK}{KD}=\frac{EJ}{JD}$ .

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#2 h Nov 11, 2015, 7:25 PM • 2 Y

Y by Adventure10, Mango247

Suppose that angle FDA=x and angle EDA=y.By Ceva, we get BF/AF*AE/CE*CD/BD=1. After using law of sines in triangles AFD and AED we get that sin(y)/sin(x)*cos(x)/cos(y)*cos(C-y)/cos(y-C)*cos(x-B)/cos(B-x)=1. We have cos(y-C)=cos(C-y) and cos(x-B)=cos(B-x). So, sin(y)/sin(x)*cos(x)/cos(y)=1. This gives us x=y and tan(x)=tan(y) which is echivalent to FK/KD=EJ/JD

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#3 h Nov 11, 2015, 8:47 PM • 2 Y

Y by UI_MathZ_25, Adventure10

Solution without trig
1. $\triangle FKP\sim \triangle CDP$ - $\frac{KF}{KP}=\frac{KF+CD}{KD}$ . So $\frac{KF}{KD}=\frac{KF}{KP}-\frac{CD}{KD}=\frac{CD}{DP}-\frac{CD}{KD}=\frac{CD\cdot KP}{KD\cdot DP}$ .
2. $\triangle EJP\sim\triangle BDP$ - $\frac{EJ}{JP}=\frac{EJ+BD}{ED}$ . So $\frac{EJ}{ED}=\frac{EJ}{JP}-\frac{BD}{ED}=\frac{BD\cdot JP}{JD\cdot DP}$ .
Therefore $\frac{KF}{KP}=\frac{EJ}{JP}$ is equivalent to
$\frac{BD\cdot JP}{JD}=\frac{CD\cdot KP}{KD}$
or
$\frac{BD}{CD}=\frac{KP}{KD}\cdot \frac{JD}{JP}$
By Ceva and obvious similarities we have
$1=\frac{BD}{CD}\cdot\frac{CE}{AE}\cdot\frac{AF}{BF}=\frac{BD}{CD}\cdot\frac{JD}{AJ}\cdot\frac{AK}{KD}$ .
Therefore
$\frac{AJ}{AK}\cdot\frac{KD}{JD}=\frac{BD}{CD}=\frac{KP}{KD}\cdot\frac{JD}{JP}$
which is equivalent to
$\frac{AJ\cdot JP}{JD^2}=\frac{AK\cdot KP}{KD^2}$ .
But
$\frac{AK\cdot KP}{KD^2}=\frac{AF}{FB}\cdot\frac{KP}{KP+CD}=\frac{AF}{FB}\cdot\frac{FP}{CF}$ .
and
$\frac{AJ\cdot JP}{JD^2}=\frac{AE}{EC}\cdot\frac{EP}{KP+BD}=\frac{AE}{EC}\cdot\frac{EP}{EB}$ .
And finally, using van Aubel's theorem and Ceva
$\frac{FB}{AF}\cdot\frac{FP+CP}{FP}=\frac{FB}{AF}\cdot\left(1+\frac{CD}{DB}+\frac{CE}{EA}\right)=\frac{FB}{AF}+\frac{CE}{EA}+\frac{FB}{AF}\cdot\frac{CE}{EA}$ ,
$\frac{EC}{EA}\cdot\frac{EP+PB}{EP}=\frac{EC}{EA}\left(1+\frac{FB}{AF}+\frac{BD}{CD}\right)=\frac{CE}{EA}+\frac{FB}{AF}+\frac{CE}{EA}\cdot\frac{FB}{AF}$ .
Hence $\frac{AK\cdot KP}{KD^2}=\frac{AJ\cdot JP}{JD^2}\Leftrightarrow \frac{FK}{KD}=\frac{EJ}{JD}$ .

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#4 h Nov 11, 2015, 10:08 PM • 3 Y

Y by Ricochet, Adventure10, Mango247

Denote $X \equiv EF \cap AD, Y \equiv EF \cap BC.$ Since the cevians $AD, BE, CF$ are concurrent in $\triangle ABC$ , we have $-1 = A(B, C; D, Y) = (F, E; X, Y).$ But since $DX \perp DY$ , it is well-known (Lemma 5) that $DX$ bisects $\angle EDF.$ Hence, $\angle EDJ = \angle FDK \implies \triangle EDJ \sim \triangle FDK$ , and the desired result follows.

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#5 h Nov 11, 2015, 11:33 PM • 3 Y

Y by alexgsi, Adventure10, Mango247

Another trig solution

First denote:

$\angle FDB= \alpha$
$\angle ADF= \beta$
$\angle EDA= \omega$
$\angle CDE= \phi$

By the generalized angle bisector theorem in $\triangle ADB \longrightarrow \frac{\sin \alpha }{\sin \beta } = \frac{BF*DA}{FA*BD}$

By the generalized angle bisector theorem in $\triangle ADB \longrightarrow \frac{\sin \omega }{\sin \phi } = \frac{AE*DC}{EC*AD}$

Apply Ceva and we get:

$\frac{\sin \alpha }{\sin \beta} * \frac{\sin \omega }{\sin \phi } = \frac{AE*DC}{EC} * \frac{BF}{FA*BD} = 1$

$\frac{\sin \alpha }{\sin \beta} = \frac{\sin \phi }{\sin \omega }$

Lemma.- Let $a+b=x+y<180$ and $\frac{\sin a}{\sin b} = \frac{\sin x}{\sin y} \longrightarrow a=x, b=y$ this can be easily done applying the generalized angle bisector theorem in an isosceles triangle.

In this case $\alpha + \beta = \phi + \omega=90$
So, we get $\alpha = \phi$ and $\beta = \omega$ which implies also using the fact that $\angle EJD = \angle FKD=90$ that $\triangle FKD \sim \triangle EJD$ and the desired result follows

This post has been edited 1 time. Last edited by Math_CYCR, Nov 11, 2015, 11:34 PM
Reason: Text

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#6 h Nov 13, 2015, 3:29 PM • 2 Y

Y by Adventure10, Mango247

Consider the homology $f: P \rightarrow P'$ that send $BC$ to the infinity, is clear $A'E'P'F'$ is a parallelogram, also $J'E' \parallel F'K' \parallel BC$ , so $F'J'E'K'$ is a parallelogram( $F'J' \parallel K'E'$ ) this implies $FJ,KE$ and $BC$ are concurrent (point $X$ )
$\Rightarrow \frac{XJ}{XF}=\frac{JD}{KD}=\frac{EJ}{FK}$

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#7 h Nov 13, 2015, 3:35 PM • 1 Y

Y by Adventure10

According to my proof we can put an easy generalization

Let $ABC$ be a triangle, and the cevians $AE,BD$ and $CF$ concurrent at $P$ , the parallel lines to $BC$ from $E$ and $F$ cut to $AD$ at $J$ and $K$ respectively. Prove that $\frac{FK}{KD}=\frac{EJ}{JD}$

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armann

#9 h Nov 13, 2015, 6:28 PM • 3 Y

Y by Adventure10, Mango247, AlexCenteno2007

fprosk wrote:

Let $ABC$ be an acute triangle and let $D$ be the foot of the perpendicular from $A$ to side $BC$ . Let $P$ be a point on segment $AD$ . Lines $BP$ and $CP$ intersect sides $AC$ and $AB$ at $E$ and $F$ , respectively. Let $J$ and $K$ be the feet of the peroendiculars from $E$ and $F$ to $AD$ , respectively. Show that

$\frac{FK}{KD}=\frac{EJ}{JD}$ .

Its very easy with blanchet theorem.

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jayme

#10 h Nov 14, 2015, 7:10 AM • 2 Y

Y by Adventure10, Mango247

Dear mathlinkers,

see also
http://www.artofproblemsolving.com/community/c6t48f6h1162621_iberoamerican_olympiad_2015_problem_4

Sincerely
Jean-Louis

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fprosk

#11 h Nov 17, 2015, 8:24 PM • 2 Y

Y by Adventure10, Mango247

This problem actually comes out rather nicely analytically. Let D be the origin, A be on the y axis, P on the y axis between D and A, and B and C on opposite sides of D on the x axis. Then you get some simple equations for the lines AB, AC, CP, and BP, which you can use to find the coordinates of E and F. FK and KD are the distances from F to the y and x axes respectively while EJ and JD are the distances from E to the y and x axes respectively, and you can see that the ratios are the same.

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#12 h Nov 17, 2015, 10:43 PM • 2 Y

Y by FabrizioFelen, Adventure10

I have a doubt
Why to select a geometry easy problem (trivial) in Iberoamerican olympiad? in recent years.

This problem I solved with projective geometry, although is trivial by Blanchet's Theorem, do to think like a region with a bad geometry

I accept all kinds of criticism, is only a doubt. Sorry my english isn't good

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jbaca

#13 h Dec 10, 2015, 6:10 PM • 3 Y

Y by drmzjoseph, Adventure10, Mango247

I agree with you. Geometry problems in the OIMs 2013, 2014, 2015 have been quite easy. I really miss a geometry problem as problem 3 or 6 in the contest. Moreover, this problem is a direct application of a very well-known theorem and an approach using projective geometry (as we can see in the previous posts) is possible. I hope that in the next OIM appear a difficult geometry problem, as in 2010 and 2011.

This post has been edited 1 time. Last edited by jbaca, Dec 10, 2015, 6:12 PM
Reason: A missing parentheses.

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#14 h Jan 10, 2016, 4:06 PM • 2 Y

Y by drmzjoseph, Adventure10

Yes you are right. It seems that the International Jury do not want choose hard geometry problems, and the results in 2014 and 2015 was that the gold medals were awarded to the students who have 42 points.
Get 41 points in an International competition and do not get god medal is, I think, very frustrating.

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