Locus of a point

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Locus of a point

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Source: APMO 1997

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3498 posts

#1 h Mar 17, 2006, 12:45 PM • 2 Y

Y by Adventure10, Mango247

Triangle $A_1 A_2 A_3$ has a right angle at $A_3$ . A sequence of points is now defined by the following iterative process, where $n$ is a positive integer. From $A_n$ ( $n \geq 3$ ), a perpendicular line is drawn to meet $A_{n-2}A_{n-1}$ at $A_{n+1}$ .
(a) Prove that if this process is continued indefinitely, then one and only one point $P$ is interior to every triangle $A_{n-2} A_{n-1} A_{n}$ , $n \geq 3$ .
(b) Let $A_1$ and $A_3$ be fixed points. By considering all possible locations of $A_2$ on the plane, find the locus of $P$ .

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484 posts

Davron

#2 h Mar 17, 2006, 1:16 PM • 2 Y

Y by Adventure10, Mango247

http://www.kalva.demon.co.uk/apmo/asoln/asol974.html

Davron

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jin

#3 h Apr 10, 2006, 12:45 PM • 2 Y

Y by Adventure10, Mango247

I don't know why all of the links of http://www.kalva.demon.co.uk I can't open.

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#4 h Feb 21, 2010, 7:23 AM • 1 Y

Y by Adventure10

Does anyone have a solution?

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#5 h Mar 4, 2016, 2:03 AM • 1 Y

Y by Adventure10

(a) It can be proved just by Sperner Theorem

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Grandmaster2000

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Grandmaster2000

#6 h Mar 4, 2016, 2:24 AM • 2 Y

Y by Adventure10, Mango247

How about power of a point?

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#7 h Mar 4, 2016, 6:02 AM • 5 Y

Y by SohamSchwarz119, JasperL, Adventure10, Mango247, Stuffybear

For a triangle $ABC$ right angled at $C$ , define its "skew-center" as the second intersection of the circles with diameters $AC$ and $BD$ , where $D$ is the foot of altitude of $ABC$ from $C$ . Note that the skew-center of $ABC$ is not necessarily the same as the skew-center of $BAC$ .

Lemma: The skew center of a right triangle lies inside the triangle.
Proof: Trivial by angle chasing.

Part a) Let $P$ be the skew-center of $A_1A_2A_3$ . We shall show that $P$ satisfies the requirements. We do so by proving that $P$ is the skew center of $A_{n-2}A_{n-1}A_{n}$ for all $n\ge 3$ . We shall use induction.

The base case $n=1$ is obvious by definition. Suppose $P$ is the skew center of $A_{k-2}A_{k-1}A_k$ . By definition, $P\in (A_{k-2}A_k)$ and $P\in (A_{k-1}A_{k+1})$ (Here $(XY)$ refers to the circle with diameter $XY$ ). Clearly $A_{k+2}\in (A_{k-1}A_{k+1})$ . Now we have $\angle A_kPA_{k+2}=2\pi-\angle A_{k+1}PA_{k+2}-\angle A_kPA_{k+1}=(\pi-\angle A_{k+1}PA_{k+2})+(\pi- \angle A_kPA_{k+1})=\angle A_{k+1}A_k+\angle A_{k+1}A_{k-1}A_{k-2}=\frac{\pi}{2}$ So $P\in (A_kA_{k+2})$ . But also $P\in (A_{k-1}A_{k+1})$ , so $P$ is the skew center of $A_{k-1}A_kA_{k+1}$ . This completes the induction, so we conclude that $P$ is the skew-center of all triangles $A_{n-2}A_{n-1}A_n$ . This together with Lemma implies that $P$ is interior to every triangle $A_{n-2} A_{n-1} A_{n}$ , $n \geq 3$ , proving our claim.

$[asy] import graph; size(7.726755742739088cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-0.7829171322827649,xmax=6.943838610456323,ymin=-0.06350701568125115,ymax=4.534562378445503; pair A_k=(0.44,1.3), P=(1.4338043821210804,1.706699099133097); D(A_k--(6.04,1.32)); D((0.4297853343707351,4.160106376194181)--(6.04,1.32)); D((0.4297853343707351,4.160106376194181)--A_k); D(A_k--(1.5904450112061543,3.5725358204778406)); D(shift((0.43489266718536757,2.7300531880970906))*xscale(1.4300623083051773)*yscale(1.4300623083051773)*arc((0,0),1,-89.79537308603418,90.20462691396585)); D(shift((3.815222505603077,2.4462679102389204))*xscale(2.49361470664756)*yscale(2.49361470664756)*arc((0,0),1,153.14971347367387,333.1497134736739)); D((1.5904450112061543,3.5725358204778406)--(1.5985464331870287,1.304137665832811)); D(shift((1.0192732165935143,1.3020688329164054))*xscale(0.5792769109262276)*yscale(0.5792769109262276)*arc((0,0),1,0.20462691396587018,180.20462691396588),linetype("2 2")); D((1.5904450112061543,3.5725358204778406)--P,linetype("2 2")); D(A_k--P,linetype("2 2")); D(P--(1.5985464331870287,1.304137665832811),linetype("2 2")); D(A_k); MP("A_k",(-0.1431857383173036,1.345901211648906),NE*lsf); D((6.04,1.32)); MP("A_{k-1}",(6.084199549815234,1.4158718328638784),NE*lsf); D((0.4297853343707351,4.160106376194181)); MP("A_{k-2}",(0.4665582465560267,4.264675696616324),NE*lsf); D((1.5904450112061543,3.5725358204778406),linewidth(3.pt)); MP("A_{k+1}",(1.6260713981184254,3.6349401056815727),NE*lsf); D(P,linewidth(3.pt)); MP("P",(1.0962938374907776,1.675762711662347),NE*lsf); D((1.5985464331870287,1.304137665832811),linewidth(3.pt)); MP("A_{k+2}",(1.0263232162758054,1.046027120727596),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

To see that this is only possible $P$ , simply observe that the sidelengths of the triangle $A_{n-2}A_{n-1}A_n$ tend to zero as $n\to\infty$ . So if there were another point $Q$ , the distance $PQ$ would eventually exceed the largest side of $A_{n-2} A_{n-1} A_{n}$ , and the triangle will not be able to accommodate both $P$ and $Q$ .

Part b) By the definition of skew-center, the required locus is clearly a subset of the circle with diameter $A_1A_3$ , excluding $A_1,A_3$ . But not all points on this circle lie on the locus.
$[asy] import graph; size(7.682604816586902cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-2.2490914484553635,xmax=5.433513368131538,ymin=-0.2146804943158581,ymax=4.357115386188629; pen qqwuqq=rgb(0.,0.39215686274509803,0.); pair A_1=(-0.46,4.3), A_3=(-0.4,0.76), A_2=(4.994373056994819,0.8514300518134715), A_4=(1.1818594352315481,3.2619215538942914), A_5=(1.2237984106033264,0.7875220069593784), A_6=(0.0761667036350339,1.5131211133793963), A_7=(1.2111740628187921,1.5323585262469175), P=(0.9311024018547345,1.398098833084234); D(arc(A_1,0.2981606009024671,-89.02897806892084,-64.38804884260522)--(-0.46,4.3)--cycle,qqwuqq); D(arc(A_3,0.2981606009024671,0.9710219310791641,25.61195115739478)--(-0.4,0.76)--cycle,qqwuqq); D(arc(A_5,0.2981606009024671,90.97102193107915,115.61195115739476)--(1.2237984106033264,0.7875220069593784)--cycle,qqwuqq); D(arc(A_2,0.2981606009024671,147.69655948789924,172.33748871421486)--(4.994373056994819,0.8514300518134715)--cycle,qqwuqq); D(A_1--A_3); D(A_1--A_2); D(A_3--A_2); D(A_4--A_5); D(A_3--A_4); D(A_5--A_6); D(A_6--A_7); D(shift((-0.43,2.53))*xscale(1.7702542190318316)*yscale(1.7702542190318316)*arc((0,0),1,-89.02897806892084,90.97102193107915),dotted); D(shift((3.0881162461131835,2.0566758028538814))*xscale(2.255312029284104)*yscale(2.255312029284104)*arc((0,0),1,147.69655948789924,327.69655948789926),dotted); D(P--A_3,linetype("2 2")); D(P--A_1,linetype("2 2")); D(P--A_5,linetype("2 2")); D(P--A_2,linetype("2 2")); D(A_1); MP("A_1",(-0.8974300576975125,3.99),NE*lsf); D(A_3); MP("A_3",(-0.8676139976072658,0.7593108019655326),NE*lsf); D(A_2); MP("A_2",(5.035965900261582,0.9481458492037614),NE*lsf); D(A_4,linewidth(3.pt)); MP("A_4",(1.2195102087100036,3.323491969726745),NE*lsf); D(A_5,linewidth(3.pt)); MP("A_5",(1.269203642193748,0.9183297891135147),NE*lsf); D(A_6,linewidth(3.pt)); MP("A_6",(-0.19178330222834047,1.6537926046729323),NE*lsf); D(A_7,linewidth(3.pt)); MP("A_7",(1.2493262688002502,1.5941604844924389),NE*lsf); D(P,linewidth(3.pt)); MP("P",(0.6231890069050695,1.3059385702867212),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$
Observe that by similarity of triangles $A_1A_2A_3$ and $A_3A_4A_5$ and by the fact that $P$ is the skew center of both of them, we have $\angle PA_1A_3=\angle PA_3A_5=\angle PA_3A_2$ . Analogously, from similar triangles $A_3A_4A_5$ and $A_5A_6A_7$ , we have $\angle PA_3A_5=\angle PA_5A_7$ . But $PA_5A_2A_4$ is cyclic, giving $\angle PA_5A_7=\angle PA_2A_1$ . Thus we conclude that $\angle PA_1A_3=\angle PA_3A_2=\angle PA_2A_1=\omega$ , the Brocard angle of $A_1A_2A_3$ . But, $\cot\omega=\cot A_3+\cot A_1+\cot A_2=0+\cot A_1+\frac{1}{\cot A_1}\ge 2\implies\tan \omega \le \frac 12\implies \omega\le \arctan {\frac 12}$ . Clearly equality can be achieved. Also $A_2$ can lie on either side of $A_1A_3$ . Thus the desired locus is an arc of $(A_1A_3)$ subtending an angle of $2\arctan {\frac 12}$ at $A_1$ and having midpoint $A_3$ , excluding the point $A_3$ .
$[asy] import graph; size(7.682604816586901cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-2.222555154975045,xmax=5.460049661611856,ymin=0.10534521731945252,ymax=4.677141097823941; pen qqwuqq=rgb(0.,0.39215686274509803,0.); pair A_1=(-0.46,4.3), A_3=(-0.4,0.76), A_2=(4.994373056994819,0.8514300518134715); D(arc(A_1,0.29816060090246704,-115.59402924599883,-62.463926891842824)--(-0.46,4.3)--cycle,qqwuqq); D(CR((-0.4,0.76),0.17638888888888887)); D(A_1--A_3); D(A_1--A_2); D(A_3--A_2); D(A_1--(1.004,1.492),linetype("2 2")); D(A_1--(-1.828,1.444),linetype("2 2")); D(shift((-0.43,2.53))*xscale(1.7702542190318316)*yscale(1.7702542190318316)*arc((0,0),1,217.84091957692317,324.10112428523513)); D(A_1); MP("A_1",(-0.9007098243074412,4.259716256560488),NE*lsf); MP("A_3",(-0.920587197700939,0.42338319161541743),NE*lsf); D(A_2); MP("A_2",(5.032686133651653,0.9501335865431084),NE*lsf); label("Locus of P",(-0.17518569544477133,1.168784693871584),NE*lsf); label("$2\arctan{\frac 12}$",(-1.0000966912749303,3.4646213208205765),NE*lsf,fp+qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

This post has been edited 10 times. Last edited by Ankoganit, Mar 5, 2016, 4:23 AM

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#9 h Mar 4, 2016, 2:17 PM • 1 Y

Y by Adventure10

Generalisation:

In triangle $A_1A_2A_3$ , angle at $A_3$ is strictly larger than other angles. A sequence of points is now defined by the following iterative process, where $n$ is a positive integer. For $n\ge 3$ , a point $A_n$ is taken on segment $A_{n-2}A_{n-3}$ such that $\angle A_{n-2}A_nA_{n-1}=\angle A_1A_3A_2$ . Prove that if this process is continued indefinitely, then one and only one point $P$ is interior to every triangle $A_{n-2} A_{n-1} A_{n}$ , $n \geq 3$ .

Also identify the point $P$ , if possible.

This post has been edited 2 times. Last edited by Ankoganit, Mar 4, 2016, 3:08 PM

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