IMO Shortlist 2010 - Problem G3

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IMO Shortlist 2010 - Problem G3

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#1 h Jul 17, 2011, 2:13 AM • 5 Y

Y by Davi-8191, anantmudgal09, Adventure10, Mango247, Funcshun840

Let $A_1A_2 \ldots A_n$ be a convex polygon. Point $P$ inside this polygon is chosen so that its projections $P_1, \ldots , P_n$ onto lines $A_1A_2, \ldots , A_nA_1$ respectively lie on the sides of the polygon. Prove that for arbitrary points $X_1, \ldots , X_n$ on sides $A_1A_2, \ldots , A_nA_1$ respectively,
$\max \left\{ \frac{X_1X_2}{P_1P_2}, \ldots, \frac{X_nX_1}{P_nP_1} \right\} \geq 1.$
Proposed by Nairi Sedrakyan, Armenia

This post has been edited 1 time. Last edited by djmathman, Nov 22, 2016, 5:20 PM

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RSM

#2 h Jul 17, 2011, 7:35 AM • 7 Y

Y by Pythagorasauras, Mobashereh, Pluto1708, amar_04, Adventure10, Mango247, Funcshun840

$P_iP_{i+1}=X_iX_{i+1}.\frac {PA_i}{2R_{\Delta A_iX_iX_{i+1}}}$ where $R_{\Delta XYZ}$ denotes the circumradius of $\Delta XYZ$ .
Claim:-
We will get some $k$ such that $P$ lies within $\odot A_kX_kX_{k+1}$ .

Proof:-
Suppose, there does not exist such $k$ . Then $\angle X_iPX_{i+1}<180-\angle X_iA_iX_{i+1}$ for all $1\le i \le k$ .
So $\sum_{i=1}^{n} \angle X_iPX_{i+1}< \sum_{i=1}^{n} 180-\angle X_iA_iX_{i+1}$
So $360<360$ which is not possible.

So our claim is proved. So $PA_k\le 2R_{A_kX_kX_{k+1}}$
So $\frac {X_kX_{k+1}}{P_kP_{k+1}}\ge 1$
So done.

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#3 h Apr 11, 2015, 6:12 PM • 4 Y

Y by Mobashereh, Mprog., Adventure10, Mango247

What an excellent problem!

Let $R(DEF)$ denote the circumdiameter of triangle $DEF$ .

Firstly notice $X_1X_2/P_1P_2 \ge 1 \Leftrightarrow X_1X_2 \ge P_1P_2 \Leftrightarrow X_1X_2/sin(\angle A_1A_2A_3) \ge P_1P_2/sin(\angle A_1A_2A_3) \Leftrightarrow R(X_1A_2X_2) \ge R(P_1A_2P_2) = A_2P$ , which happens iff $P$ lies inside circle $X_1A_2X_2$ , which happens iff $\angle X_1PX_2 \ge 180-\angle A_1A_2A_3$ .

Now if this were false always then add all the $n$ equations, and we get that $360 = \sum \angle X_iPX_{i+1} < \sum (180 - \angle A_{i-1}A_iA_{i+1}) = 360$ , impossible!

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va2010

#4 h Oct 12, 2015, 11:43 AM • 1 Y

Y by Adventure10

The problem statement is wrong as stated. Could someone correct it? Thank you.

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#6 h Nov 22, 2016, 5:18 PM • 1 Y

Y by Adventure10

va2010 wrote:

The problem statement is wrong as stated. Could someone correct it? Thank you.

EDIT: should be $P_1P_2$

Fixed ~dj

This post has been edited 3 times. Last edited by djmathman, Nov 22, 2016, 5:20 PM

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#7 h Jun 11, 2017, 7:17 AM • 5 Y

Y by Pluto1708, amar_04, Adventure10, Mango247, Funcshun840

We should see this type of problems more often in math olympiads

IMO ShortList 2010 G3 wrote:

Let $A_1A_2 \ldots A_n$ be a convex polygon. Point $P$ inside this polygon is chosen so that its projections $P_1, \ldots , P_n$ onto lines $A_1A_2, \ldots , A_nA_1$ respectively lie on the sides of the polygon. Prove that for arbitrary points $X_1, \ldots , X_n$ on sides $A_1A_2, \ldots , A_nA_1$ respectively,
$\max \left\{ \frac{X_1X_2}{P_1P_2}, \ldots, \frac{X_nX_1}{P_nP_1} \right\} \geq 1.$
Proposed by Nairi Sedrakyan, Armenia

Notice that $\sum^n_{i=1} \measuredangle X_iA_iX_{i+1}+\sum^n_{i=1} \measuredangle X_iPX_{i+1}=(n-2)\pi+2\pi=n\pi,$ with indices mod $n$ , so by pigeon-hole principle, there exists an index $i$ for which $\measuredangle X_iA_iX_{+1}+\measuredangle X_iPX_{i+1} \geqslant \pi$ .

Claim: $X_iX_{i+1} \geqslant P_iP_{i+1}$ .

(Proof) Let $X$ be the antipodes of $A$ in $\triangle X_iAX_{i+1}$ ; then, $\measuredangle X_iAX_{i+1}+ \measuredangle X_iPX_{i+1} \geqslant \pi \implies AX \cdot \, \sin X_iAX_{i+1} \geqslant AP\cdot \, \sin P_iAP_{i+1} \implies X_iX_{i+1} \geqslant P_iP_{i+1}. \, \blacksquare$
Hence, the statement of the initial problem is proved. $\blacksquare$

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#9 h Mar 25, 2020, 9:08 PM

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Note that if $X_iX_{i+1}<P_iP_{i+1}$ , then the circle $(X_iA_{i+1}X_{i+1})$ has a smaller diameter than $(P_iA_{i+1}P_{i+1})$ , and therefore does not contain $P$ . Hence, $\angle X_iPX_{i+1}<\angle P_iPP_{i+1}$ . However, since $\sum \angle X_iPX_{i+1}=\sum\angle P_iPP_{i+1}=360^o$ , this cannot hold for all values of $i$ , which gives the desired result.

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#10 h May 10, 2020, 10:46 AM

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Here is a different solution:

For points $U,V$ on directed segment $\stackrel{\longrightarrow}{AB}$ , we say $U$ is on the left side of $V$ if $A,U,V,B$ are aligned in that order, and $U$ is on the right side of $V$ if $A,V,U,B$ are in order. All indices are taken modulo $n$ , and direct all segments in $A_iA_{i+1}$ from $A_i$ to $A_{i+1}$ .

First notice that $\angle A_iP_{i-1}P_i , \angle A_iP_iP_{i-1}$ are both acute because of the existence of $P$ . If there exists an index $i$ , such that $X_i$ and $X_{i+1}$ are respectively on the left and right side of $P_i$ and $P_{i+1}$ , then we can prove that $X_iX_{i+1}>P_iP_{i+1}$ . (Draw a line through $X_i$ parallel to $P_iP_{i+1}$ and let this line cut $A_{i+1}A_{i+2}$ at point $X'$ between $P_{i+1}$ and $X_{i+1}$ . Then $X_iX_{i+1} \ge X_iX' > P_iP_{i+1}$ .) Therefore, we can WLOG suppose all $X_i$ lie on the right side of $P_i$ .

Let $P_iX_i = d_i$ . We claim that $\frac{d_i}{PP_i} > \frac{d_{i+1}}{PP_{i+1}}$ . Take $i=1$ for example: fix $X_1$ , and suppose another point $X_2'$ lies on the right side of $P_2$ and satisfies $\frac{P_1X_1}{PP_1} > \frac{P_2X_2'}{PP_{2}}$ . If $X_2$ does not lie on the left side of $X_2'$ , then because $\measuredangle(X_1X_2',A_2A_3) > \frac{\pi}{2}$ , $X_1X_2 \ge X_1X_2' = P_1P_2 \cdot \frac{P_2X_2'}{PP_2} > P_1P_2.$ So $X_2$ lies between $P_2$ and $X_2'$ , which proves the claim.

Now $\frac{d_1}{PP_1} > \frac{d_2}{PP_2} > \dots >\frac{d_n}{PP_n} > \frac{d_1}{PP_1}$ , a contradiction.

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anser

#11 h Dec 7, 2020, 5:02 PM

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Incorrect

@below

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#12 h Dec 7, 2020, 5:52 PM

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@anser

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#13 h Aug 9, 2021, 8:12 AM

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Two different solutions: (though, not as elegant as some previous one)

Solution 1:

Whenever we would write segment in our solution, we will also consider the two endpoints of the segment (unless specified).
We consider two cases:

CASE 1: There exist a index $i$ such that both of $P_{i},P_{i+1}$ do not lie on segments $\overline{A_{i+1}P_{i}},\overline{A_{i+1}P_{i+1}}$ , respectively.
$[asy] pair A2=dir(90),P=dir(-90),P1=dir(-150),P2=dir(-40),X1=1.5*P1-0.5*A2,X2=1.8*P2-0.8*A2,T=foot(P,P1,P2),T1=foot(X1,P,T),Y1=extension(X1,T1,A2,X2),Y2=extension(X2,foot(X2,P,T),A2,X1); dot("$A_2$",A2,dir(A2)); dot("$P$",P,dir(P)); dot("$P_1$",P1,dir(-180)); dot("$P_2$",P2); dot("$X_1$",X1,dir(-180)); dot("$X_2$",X2,dir(0)); dot("$Y_1$",Y1,dir(0)); dot("$Y_2$",Y2,dir(-180)); draw(P1--P2^^X1--Y1^^X2--Y2,red); draw(Y2--A2--X2,green); draw(P2--P--P1,purple); D(X1--X2); markscalefactor=0.03; draw(rightanglemark(P,P2,A2),darkblue); draw(rightanglemark(P,P1,A2),darkblue); [/asy]$
We will prove that $X_{i}X_{i+1} \ge P_{i}P_{i+1}$ . For the sake of simplicity, let $i=1$ . The main thing to note is that both the angles $\angle AP_1P_2, \angle AP_2P_1$ are acute. Now, let $Y_1,Y_2$ be points on lines $\overline{AP_2},\overline{AP_1}$ such that $\overline{P_1P_2} \parallel \overline{X_1Y_1} \parallel \overline{X_2Y_2}$ . Note that it cannot happen that $Y_1,Y_2$ both do not lie on segments $\overline{AX_1},\overline{AX_2}$ , respectively (otherwise, the parallel lines $\overline{X_1Y_1},\overline{X_2Y_2}$ would intersect). So WLOG $Y_1$ does not lie on segment $\overline{AX_2}$ . Then, $\angle X_1Y_1X_2 = 180^\circ - \angle AP_2P_1 \ge 90^\circ$ . Hence, $X_1X_2 \ge X_1Y_1 \ge P_1P_2$ . $\square$

CASE 2: WLOG assume $X_1$ lies on segment $\overline{A_2P_1}$ . Then for all $i$ , we have that $X_i$ lies on segment $\overline{P_iA_{i+1}}$ .

Then at least one of the following inequalities must hold:
$\angle X_1PP_1 \le \angle X_2PP_2 ~,~ \angle X_2PP_2 \le \angle X_3PP_3 ~, ~ \ldots ~,~ \angle X_nPP_n \le \angle X_1PP_1$ WLOG, suppose $\angle X_1PP_1 \le \angle X_2PP_2$ . We will prove that $X_1X_2 \ge P_1P_2$ .
$[asy] pair A2=dir(90),P=dir(-90),P1=dir(-150),P2=dir(-40),X1=0.7*P1+0.3*A2,X2=1.45*P2-0.45*A2,X2p=2*foot(circumcenter(A2,X1,P),A2,P2)-A2; path c1=circumcircle(A2,P1,P2),c2=circumcircle(A2,X1,P); dot("$A_2$",A2,dir(A2)); dot("$P$",P,dir(P)); dot("$P_1$",P1,dir(-180)); dot("$P_2$",P2); dot("$X_1$",X1,dir(-180)); dot("$X_2$",X2,dir(-50)); dot("$X_2'$",X2p,dir(-30)); draw(c1^^c2,red); draw(P1--A2--X2,green); draw(P1--P--X1^^P2--P--X2p^^X2--P,purple); D(X2p--X1--X2); D(P1--P2); markscalefactor=0.02; draw(rightanglemark(P,P2,A2),darkblue); draw(rightanglemark(P,P1,A2),darkblue); markscalefactor=0.03; filldraw(anglemark(X1,P,P1),fuchsia); filldraw(anglemark(X2p,P,P2),fuchsia); [/asy]$
Let $X_2'$ be a point on segment $\overline{X_2P_2}$ such that $\angle X_1PP_1 = \angle X_2'PP_2$ . As projection of $X_1$ onto line $\overline{A_2P_2}$ does not lie on segment $\overline{X_2X_2'}$ , so $\angle X_1X_2'X_2 \ge 90^\circ$ implying $X_1X_2 \ge X_1X_2$ and hence it suffices to show $X_1X_2' \ge P_1P_2$ . Note that both the quadrilateral $AP_1PP_2$ and $AX_1PX_2'$ are cyclic so by Ptolemy's theorem,
$\begin{align*} PA_2 \cdot X_1X_2' &= PX_1 \cdot AX_2' + PX_2' \cdot A_2X_1 \ge PP_1 \cdot AX_2' + PP_2 \cdot A_2X_1 \\ &= (PP_1 \cdot P_2X_2' - PP_2 \cdot P_1X_1) + PP_1 \cdot A_2P_2 + PP_2 \cdot A_2P_1 = PA_2 \cdot P_1P_2 \end{align*}$ where the equality $PP_1 \cdot P_2X_2' = PP_2 \cdot P_1X_1$ follows from $\triangle PP_1X_1 \sim \triangle PP_2X_2'$ . $\square$

This completes the proof of the problem. $\blacksquare$

Solution 2:

Whenever we would write segment in our solution, we will also consider the two endpoints of the segment (unless specified).
Following is the key lemma:

Lemma: Given a triangle $LMN$ with $\angle M , \angle N \le 90^\circ$ let $R,S$ be any two points on lines $\overline{LM},\overline{LN}$ such that points $M,N$ lie on segments $\overline{LR},\overline{LS}$ , respectively. Then $RS \ge MN$ .
$[asy] pair A2=dir(90),P=dir(-90),P1=dir(-150),P2=dir(-40),X1=1.5*P1-0.5*A2,X2=1.8*P2-0.8*A2,T=foot(P,P1,P2),T1=foot(X1,P,T),Y1=extension(X1,T1,A2,X2),Y2=extension(X2,foot(X2,P,T),A2,X1); dot("$L$",A2,dir(A2)); dot("$M$",P1,dir(-180)); dot("$N$",P2); dot("$R$",X1,dir(-180)); dot("$S$",X2,dir(0)); dot("$R_0$",Y1,dir(0)); dot("$S_0$",Y2,dir(-180)); draw(P1--P2^^X1--Y1^^X2--Y2,red); draw(Y2--A2--X2,green); D(X1--X2); [/asy]$
proof: Let $R_0,S_0$ be points on lines $\overline{LN},\overline{LM}$ such that $\overline{MN} \parallel \overline{RR_0} \parallel \overline{SS_0}$ . At least (in fact exactly) one of $R_0,S_0$ must lie on segments $\overline{NS},\overline{MR}$ , respectively (otherwise, the two parallel lines $\overline{RR_0},\overline{SS_0}$ would intersect). WLOG, $R_0$ lies on segment $NS$ . Now, $\angle RR_0S = \angle MNS \ge 90^\circ$ and thus
$RS \ge RR_0 \ge MN$ This proves our Lemma. $\square$

Now if there would be a index $i$ such that both of $P_{i},P_{i+1}$ do not lie on segments $\overline{A_{i+1}P_{i}},\overline{A_{i+1}P_{i+1}}$ , respectively ; then our Lemma would directly give $X_{i}X_{i+1} \ge P_{i}P_{i+1}$ .

Now WLOG assume $X_1$ lies on segment $\overline{A_2P_1}$ and the consider the case when $X_i$ lies on segment $\overline{P_iA_{i+1}}$ for all $i$ .

Then at least one of the following inequalities must hold:
$\angle X_1PP_1 \le \angle X_2PP_2 ~,~ \angle X_2PP_2 \le \angle X_3PP_3 ~, ~ \ldots ~,~ \angle X_nPP_n \le \angle X_1PP_1$ WLOG, suppose $\angle X_1PP_1 \le \angle X_2PP_2$ . We will prove that $X_1X_2 \ge P_1P_2$
$[asy] pair A2=dir(90),P=dir(-90),P1=dir(-150),P2=dir(-40),X1=0.7*P1+0.3*A2,X2=1.45*P2-0.45*A2,X2p=2*foot(circumcenter(A2,X1,P),A2,P2)-A2; path c1=circumcircle(A2,P1,P2),c2=circumcircle(A2,X1,P); dot("$A_2$",A2,dir(A2)); dot("$P$",P,dir(P)); dot("$P_1$",P1,dir(-180)); dot("$P_2$",P2); dot("$X_1$",X1,dir(-180)); dot("$X_2$",X2,dir(-50)); dot("$X_2'$",X2p,dir(-30)); draw(c1^^c2,red); draw(P1--A2--X2,green); draw(P1--P--X1^^P2--P--X2p^^X2--P,purple); D(X2p--X1--X2); D(P1--P2); markscalefactor=0.02; draw(rightanglemark(P,P2,A2),darkblue); draw(rightanglemark(P,P1,A2),darkblue); markscalefactor=0.03; filldraw(anglemark(X1,P,P1),fuchsia); filldraw(anglemark(X2p,P,P2),fuchsia); [/asy]$
Let $X_2'$ be a point on segment $\overline{X_2P_2}$ such that $\angle X_1PP_1 = \angle X_2'PP_2$ . As projection of $X_1$ onto line $\overline{A_2P_2}$ does not lie on segment $\overline{X_2X_2'}$ , so $\angle X_1X_2'X_2 \ge 90^\circ$ implying $X_1X_2 \ge X_1X_2$ and hence it suffices to show $X_1X_2' \ge P_1P_2$ .

Consider the rotation $\mathcal R$ at $P$ with angle $X_1PP_1 = \angle X_2'PP_2$ in anti-clockwise direction. Since $PX_1 \ge PP_1$ and $PX_2' \ge PP_2$ ; so $\mathcal R$ will map $X_1,X_2'$ to two points lying on lines $\overline{PP_1},\overline{PP_2}$ beyond $P_1,P_2$ (or equal to $P_1$ or $P_2$ ), respectively. Now since $\angle P_1,\angle P_2 \le 90^\circ$ in $\triangle PP_1P_2$ and since rotation preserves lengths of segments, so we are again done by our Lemma.

This completes the proof of the problem. $\blacksquare$

This post has been edited 1 time. Last edited by guptaamitu1, Aug 9, 2021, 8:13 AM
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#14 h Jan 21, 2022, 5:29 PM • 1 Y

Y by MerlinLegion07

HINT

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awesomeming327.

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awesomeming327.

#15 h May 23, 2023, 4:02 AM

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Remarkable.

Pick $X_n$ , $X_{n+1}$ such that $\angle X_nPX_{n+1} \le \angle P_nPP_{n+1}$ , which must exist. Then, $P$ lies inside of the circle $(AX_nX_{n+1})$ . Let its radius be $r$ . Note that
$\frac{X_nX_{n+1}}{P_nP_{n+1}}=\frac{2r\sin(\angle P_nA_nP_{n+1})}{PA_n\sin(\angle P_nA_nP_{n+1})}=\frac{2r}{PA_n}\ge 1$ since $P$ lies inside the circle.

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#16 h Nov 11, 2023, 11:33 PM

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pain

hello,
The circumdiameter of $A_iX_{i-1}X_i$ is always less than that of $A_iP_{i-1}P_i$ , which is $A_iP$ . Now since the smallest circle passing through $A_i$ which contains $P$ has diameter at least $A_iP$ it follows that $P$ lies outside $A_iX_{i-1}X_i$ 's circumcircle, which is not possible since it implies $\measuredangle X_{i-1}PX_i<\measuredangle P_{i-1}PP_i$ or by summing $360^{\circ}<360^{\circ}$ .

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#17 h Jan 11, 2024, 11:13 PM

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$[asy] unitsize(2cm); pair Ai,Ai1,Ai2,P,Pi,Pi1,Xi,Xi1; Ai=(0,0); Ai1=(3,1.5); Ai2=(7,0); P=(2.5,-2); Pi=(1.2,0.6); Pi1=(3.7123287671233,1.2328767123288); Xi=(2.3336071475857,1.1668035737929); Xi1=(4.5524681642484,0.9178244384069); draw(circle((2.75,-0.25),sqrt(3.125))); draw(circle((3.3274402002021,0.0121285340779),sqrt(2.3209785838135))); draw(Ai--Ai1--Ai2--cycle); draw(P--Pi--Pi1--cycle); draw(Xi--Xi1); label("$P$",P,S); label("$P_i$",Pi,NW); label("$P_{i+1}$",Pi1,N); label("$A_i$",Ai,W); label("$A_{i+1}$",Ai1,N); label("$A_{i+2}$",Ai2,E); label("$X_i$",Xi,NW); label("$X_{i+1}$",Xi1,NE); [/asy]$

Suppose by contradiction that there exist some configuration of the points $X_1,X_2,\cdots,X_n$ chosen respectively on the sides $A_1A_2, A_2A_3,\cdots , A_nA_1$ such that $\frac{X_iX_{i+1}}{P_iP_{i+1}}<1$ and thus $X_iX_{i+1}<P_iP_{i+1}$ for all $i$ , where indices are taken modulo $n$ .

Since $\angle PP_iA_{i+1}=\angle PP_{i+1}A_{i+1}=90^\circ$ by construction, we find that the quadrilateral $PP_iA_{i+1}P_{i+1}$ must always be cyclic, so that the circumcircle of $\triangle P_iA_{i+1}P_{i+1}$ has radius $\frac{PA_{i+1}}2$ . Hence, using the Law of Sines, we find that $\frac{P_iP_{i+1}}{\sin\angle P_iA_{i+1}P_{i+1}}=PA_{i+1}$ and thus $P_iP_{i+1}=PA_{i+1}\sin\angle A_{i+1}.$ Similarly, applying the Law of Sines on $\triangle X_iA_{i+1}X_{i+1}$ gives $X_iX_{i+1}=2R_i\sin\angle A_{i+1},$ where $R_i$ is the radius of the circumcircle of $\triangle X_iA_{i+1}X_{i+1}$ , so that our initial assumption gives
$\begin{align*} 2R_i&=\frac{X_iX_{i+1}}{\sin\angle A_{i+1}}\\ &<\frac{P_iP_{i+1}}{\sin\angle A_{i+1}}\\ &=PA_{i+1} \end{align*}$ for all $i$ . In particular, it follows that $P$ must be strictly outside the circumcircle $(X_iA_{i+1}X_{i+1})$ . The points $P$ and $A_{i+1}$ are on opposite sides of the segment $X_iX_{i+1}$ , so it follows for all $i$ that $\angle X_iA_{i+1}X_{i+1}+\angle X_iPX_{i+1}<180^\circ.$
Summing the above inequality then yields
$\begin{align*} 180n^\circ&>(\angle X_1A_2X_2+\angle X_2A_3X_3+\cdots+\angle X_nA_1X_1)+(\angle X_1PX_2+\angle X_2PX_3+\cdots+\angle X_nPX_1)\\ &=(\angle A_1A_2A_3+\angle A_2A_3A_4+\cdots+\angle A_nA_1A_2)+(\angle X_1PX_2+\angle X_2PX_3+\cdots+\angle X_nPX_1)\\ &=180(n-2)^\circ+360^\circ\\ &=180n^\circ, \end{align*}$ a contradiction. Hence our original assumption was wrong, thus completing the proof. $\blacksquare$

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#18 h Jan 9, 2025, 3:53 AM • 1 Y

Y by dolphinday

This is really cute.

Claim: There exists an index $i$ such that $\angle X_i PX_{i+1} \geq 180^\circ - \angle X_i A_{i+1}X_{i+1}$ , where indices are taken modulo $n$ .

Proof: Notice that for $\alpha_i = \angle X_i P X_{i+1}$ and $\beta_i = 180^\circ - \angle X_i A_{i+1} X_{i+1}$ for each $i$ , we have $\sum_{i=1}^n \alpha_i = 360^\circ = \sum_{i=1}^n \beta_i$ . Thus by Pigeonhole, there exists an $i$ with $\alpha_i \geq \beta_i$ , as needed. $\blacksquare$

For this $i$ , it follows that $P$ lies inside $(X_iA_{i+1}X_{i+1})$ as $P$ and $A_{i+1}$ are on opposite sides of $\overline{X_i X_{i+1}}$ . Thus $PA_{i+1} \leq 2R_{(X_iA_{i+1}X_{i+1})} = \frac{X_iX_{i+1}}{\sin \beta_i}$ . But on the other hand, $P_i$ and $P_{i+1}$ lie on $(PA_{i+1})$ , so $P_iP_{i+1} = PA_{i+1} \sin \beta_i \leq X_iX_{i+1}$ , as needed.

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