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
Problem
Source:
Tags: geometry, circumcircle, Inequality, polygon, IMO Shortlist
17.07.2011 10:35
$ 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.
11.04.2015 21:12
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!
12.10.2015 14:43
The problem statement is wrong as stated. Could someone correct it? Thank you.
22.11.2016 20:18
va2010 wrote: The problem statement is wrong as stated. Could someone correct it? Thank you. EDIT: should be $P_1P_2$ Fixed ~dj
11.06.2017 10:17
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. $\blacksquare$
26.03.2020 00:08
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.
10.05.2020 13:46
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.
07.12.2020 20:02
07.12.2020 20:52
09.08.2021 11:12
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][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][/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][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][/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][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][/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][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][/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. $\blacksquare$
21.01.2022 20:29
23.05.2023 07:02
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.
12.11.2023 02:33
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 $360^{\circ}<360^{\circ}$.
12.01.2024 02:13
[asy][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][/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. $\blacksquare$
09.01.2025 06:53
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.