x,y,z collinear

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

x,y,z collinear

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Iran TST 2015,second exam,second day,problem 4

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

228 posts

ATimo

#1 h Jun 4, 2015, 11:31 AM • 3 Y

Y by buratinogigle, AdithyaBhaskar, Adventure10

Let $\triangle ABC$ be an acute triangle. Point $Z$ is on $A$ altitude and points $X$ and $Y$ are on the $B$ and $C$ altitudes out of the triangle respectively, such that:
$\angle AYB=\angle BZC=\angle CXA=90$
Prove that $X$ , $Y$ and $Z$ are collinear, if and only if the length of the tangent drawn from $A$ to the nine point circle of $\triangle ABC$ is equal with the sum of the lengths of the tangents drawn from $B$ and $C$ to the nine point circle of $\triangle ABC$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

27 posts

#2 h Jun 4, 2015, 7:34 PM • 1 Y

Y by Adventure10

seems nice... Does anyone have a solution?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

824 posts

andria

#3 h Jun 12, 2015, 10:20 AM • 6 Y

Y by Rmasters, buratinogigle, AdithyaBhaskar, Chokechoke, Adventure10, Mango247

My solution:
Let $E,F,D$ feet of $B,C,A$ on $AC,AB,BC$ :
First assume that $X,Y,Z$ are collinear then let $XC\cap YB=T$ in triangle $CZB$ : $CZ^2=CD.CB$ (1) and in triangle $CXA$ : $CE.CA=CX^2$ (2) because $CE.CA=CD.CB$ from (1),(2) we get that $CZ=CX$ similarly we get that $BZ=BY$ note that $AX^2=AE.AC=AF.AB=AY^2\longrightarrow AX=AY$ because $X,Y,Z$ are collinear $\angle YZB+\angle CZX+90=180\longrightarrow \angle YZB+\angle CZX=90$ * so because $\angle BZY=\angle BYZ,\angle CZX=\angle CXZ$ from * : $\angle TYX+\angle TXY=90\longrightarrow \angle XTY=90$ so $AXTY$ is square( $AX=AY$ ) and $ZBTC$ is rectangle $\longrightarrow TC=BZ=BY\longrightarrow BY+CX=AX$ ** note that the lengths of tangents from $A,B,C$ WRT nine point circle are equal $x=\sqrt{AE.\frac{AC}{2}},y=\sqrt{BF.\frac{AB}{2}},z=\sqrt{CE.\frac{AC}{2}}$ but observe that $AE.AC=AX^2,BF.AB=BY^2,CE.AC=CX^2$ so $x=\frac{AX}{\sqrt{2}},y=\frac{BY}{\sqrt{2}},z=\frac{CX}{\sqrt{2}}$ but according to ** $y+z=x$ .
You can prove the inverse in a similar way.
DONE

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

445 posts

#4 h Jun 12, 2015, 7:23 PM • 3 Y

Y by TelvCohl, Adventure10, Mango247

If $X,Y,Z$ are collinear is trivial solution, the part hard of the problem is when $YB+XC=AY$ (is equivalent to "the length of the tangent drawn from $A$ to the nine point circle of $\triangle ABC$ is equal with the sum of the lengths of the tangents drawn from $B$ and $C$ to the nine point circle of $\triangle ABC$ ")

$\gamma$ is the semiplane that determine $BC$ and where lies on $A$
If we fixed $\tau_1$ and $\tau_2$ the circles of centers $B,C$ and radius $YB,CX$ , then only exist a point $P$ on $\gamma$ and on the radical axis of $\tau_1$ and $\tau_2$ such that the common power of $P$ is $(YB+XC)^2$ , and this is $A$ .
The externally bisector angle of $\angle BZC$ cut again to $\tau_1$ and $\tau_2$ at $Y'$ and $X'$ the tangents to $\tau_1$ and $\tau_2$ from $Y'$ and $X'$ cut at $A'$ and since $\angle Y'A'X' =\angle Y'BZ=\angle X'CZ=90^{\circ} \wedge \angle BYZ=\angle CXZ=45^{\circ}$ is easy that $A'Y'=BZ+ZC \Rightarrow A' \equiv A, Y' \equiv Y, X' \equiv X$ is sufficient.

This post has been edited 1 time. Last edited by drmzjoseph, Jun 12, 2015, 8:26 PM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a