AoPS - Problem

Problem

Source: IMO ShortList 2001, geometry problem 7

Tags: geometry, circumcircle, perimeter, perpendicular, IMO Shortlist

orl

30.09.2004 20:45

Let $O$ be an interior point of acute triangle $ABC$. Let $A_1$ lie on $BC$ with $OA_1$ perpendicular to $BC$. Define $B_1$ on $CA$ and $C_1$ on $AB$ similarly. Prove that $O$ is the circumcenter of $ABC$ if and only if the perimeter of $A_1B_1C_1$ is not less than any one of the perimeters of $AB_1C_1, BC_1A_1$, and $CA_1B_1$.

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orl

30.09.2004 20:45

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

pohoatza

26.05.2007 19:03

Let us denote by $p_{XYZ}$ the perimeter of a triangle $XYZ$. "$\Rightarrow$" If $O$ is the circumcenter of $\triangle{ABC}$, then $A_{1},B_{1}, C_{1}$ are the midpoints of the sides of the triangle, therefore $p_{A_{1}B_{1}C_{1}}=p_{AB_{1}C_{1}}= p_{A_{1}BC_{1}}= p_{A_{1}B_{1}C}$. "$\Leftarrow$" Assume WLOG, that $p_{A_{1}B_{1}C_{1}}$ is greater than $p_{AB_{1}C_{1}}$, $p_{A_{1}BC_{1}}$, $p_{A_{1}B_{1}C}$. And let $\alpha_{1}$, $\alpha_{2}$, $\beta_{1}$, $\beta_{2}$, $\gamma_{1}$, $\gamma_{2}$ be $\angle{B_{1}A_{1}C}$, $\angle{C_{1}A_{1}B}$, $\angle{C_{1}B_{1}A}$, $\angle{A_{1}B_{1}C}$, $\angle{A_{1}C_{1}B}$, $\angle{B_{1}C_{1}A}$. Now suppose $\gamma_{1}, \beta_{2}\geq \alpha$. If $A_{2}$ is the fourth vertex of the parallelogram $B_{1}AC_{1}A_{2}$, then these conditions imply that $A_{1}$ is in the interior or on the sides of $\triangle{B_{1}C_{1}A_{2}}$, therefore $p_{A_{1}B_{1}C_{1}}\leq p_{A_{2}B_{1}C_{1}}\leq p_{AB_{1}C_{1}}$. But, lets see that if one of the inequalitites $\gamma_{1}\geq \alpha$ and $\beta_{2}\geq \alpha$ is strict, then $p_{A_{1}B_{1}C_{1}}< p_{AB_{1}C_{1}}$, which contradicts our assumption. Therefore $\beta_{2}\geq \alpha$ implies $\gamma_{1}\leq \alpha$, and the analogues. $(\star)$ Now, because of the symmetry, we can suppose WLOG, that $\gamma_{1}\leq \alpha$, then since $\gamma_{1}+\alpha_{2}=180-\beta=\alpha+\gamma$, it follows that $\alpha_{2}\geq \gamma$. From $(\star)$ we deduce that $\beta_{1}\leq \gamma$, thus $\gamma_{2}\geq \beta$, $\alpha_{1}\leq \beta$, and $\beta_{2}\geq \alpha$. Now putting them all together, we obtain that: $\gamma_{1}\leq \alpha\leq \beta_{2}$, $\alpha_{1}\leq \beta\leq \gamma_{2}$, $\beta_{1}\leq \gamma\leq \alpha_{2}$. Now since $OA_{1}BC_{1}$ and $OB_{1}CA_{1}$ are cyclic quadrilaterals, we have that $\angle{A_{1}OB}=\gamma_{1}$ and $\angle{A_{1}OC}=\beta_{2}$. Thus $\frac{BO}{CO}=\frac{\cos{\beta_{2}}}{\cos{\gamma_{1}}}$, so $BO\leq CO$. Now in a similar way prove that $CO \leq AO$ and $AO \leq BO$. Thus $OA=OB=OC$, therefore $O$ is the circumcenter.

estoyanovvd

23.11.2010 15:37

It is my proposal for IMO 2001 !

GoldenFrog1618

04.02.2011 04:39

Let $P$ be the circumcenter of $ABC$. Let $D,E,F$ be the midpoints of $BC,AC,AB$, respectively and $G$ the foot of $A$ to $BC$. Obviously if $O=P$, then the perimeters of all four triangles are equal. Lemma 1: If $O\in AP, O\neq P$ then the perimeter $BA'C'$ is greater than the perimeter of $A'B'C'$. Proof: It is easily proven that $B'C'\|EF\|BC$. Draw a line $l$ parallel to $A'B'$ such that $l$ passes through $C'$. Let $Q=l\cap BC$. We have that $Q$ is between $B$ and $A'$ because, \[A'B\ge BD>B'C'\] if $\angle C\ge \angle B$ and \[A'B=DB-DA'=DB-DG+DG(AO/AP)=DB-DG+DG(B'C'/FE)>DB(B'C'/FE)=B'C'\] if $\angle B\ge \angle C$. By the triangle inequality $BC'+BQ>C'Q=A'B'$ so $BC'+BQ+B'C'=BC'+BA'>A'B'+B'C'$, as desired. $\boxed{}$. Lemma 2: If $O\in APE$ with feet $A',B',C'$ and $O'\in AP$ with feet $A''B'C''$. Then the $BA''+BC''-B'A''-B'C''> BA'+BC'-B'A'-B'C'$. Proof: $BA''>BA'$ and $BC''>BC'$, obviously. We also have that $-B'A''>-B'A'$ and $-B'C''>-B'C'$ because $90^{\circ}>\angle AC''B'>\angle AC'B'$ and$90^{\circ}>\angle CA''B'>\angle CA'B'$ . Adding the four inequalities gives the desired. $\boxed{}$. Thus, for any point $O$ in without loss of generality $APC$, we have that the perimeter of $A'B'C'$ is less than the perimeter of $BC'A'$, as desired.

Zhero

27.02.2011 00:15

Here's a much longer solution... I'm not certain that this is correct.

Let $R$ be the circumradius of $\triangle ABC$. Lemma 1: $B_1C_1 = \frac{OA \cdot BC}{2R}$. Proof: $AC_1OB_1$ is cyclic, and has diameter $AO$, so by the law of sines $\frac{B_1C_1}{\sin A} = OA$, so $B_1C_1 = OA \sin A = \frac{OA \cdot BC}{2R}$. Lemma 2: $2R \geq \max(AB, BC, CA)$. Proof: This follows trivially from the fact that the length of a longest possible chord in a circle is the diameter. Lemma 3: Let $P$ be any point not in the exterior of $\triangle ABC$. Then $2R \geq PA + PB + PC - \max(PA, PB, PC)$. Proof: Let $f_1(P) = PB+PC$, $f_2(P) = PC+PA$, $f_3(P) = PA+PB$, and $f(P) = \min(f_1(P), f_2(P), f_3(P))$. We wish to show that $f(P) \leq 2R$. We will first prove this when $P$ is on the perimeter of $\triangle ABC$. Without loss of generality, let it be on side $AB$. Then $f(P) = PA + PB + PC - \max(PA,PB,PC) \leq PA + PB = AB \leq 2R$ Suppose now that $O'$ is the circumcenter of $\triangle ABC$. If $P = O'$, then we are immediately done. Otherwise, without loss of generality suppose that $O'$ is the origin, let $u$ be the vector from $O'$ to $P$, and for any nonnegative real $t$ let $\alpha(t) = tu$. (The image of $\alpha$ would be the ray $\overrightarrow{O'P}$.) Let $g_i(t) = f_i(\alpha(t))$ for $i = 1,2,3$ and $g(t) = \min(g_1(t), g_2(t), g_3(t))$. We may suppose without loss of generality that $g_1(1) = g(1)$. We claim that $g_1(t) = g(t)$ for all nonnegative $t$. If $g_1(t) \neq g(t)$ for some $t$, then $g_j(t) < g_1(t)$ for some $j$, which we may suppose without loss of generality equals 2. Because $g_2(1) \geq g_1(1)$, the intermediate value theorem implies the existence of some $c \in [\min(t, 1), \max(t, 1)]$ such that $g_1(c) = g_2(c)$. Noting that $t \neq 0$ (since $g_1(0) = g_2(0)$ while $g_1(t) \neq g_2(t)$), we find that $c \geq \min(t, 1) > 0$. Letting $X = \alpha(c)$, we find that $BX + CX = f_1(X) = f_2(X) = AX + CX$, so $AX = BX$. It follows that $X$ must be equidistant from $A$ and $B$, so $X$ must lie on the perpendicular bisector of $AB$. Because $X$ also lies on the line through $\alpha(t)$ and $O$, we find that either $X = O$ or $\alpha(t)$ lies on the perpendicular bisector of $AB$. The former case is impossible since $c \neq 0$. In the latter case, we find that every point on the image of $\alpha$ is equidistant from $A$ and $B$, so $g_1(t) = g_2(t) < g_1(t)$, which is impossible. Hence, it is safe for us to assume that $g(t) = g_1(t)$ for all nonnegative $t$. We now observe that $g = g_1$ is convex for all nonnegative $t$; this can easily be verified from the fact that as $P$ varies along $\overrightarrow{O'P}$, the distances $PB$ and $PC$ are convex in the $x$-coordinate of $P$ (which can easily be verified with coordinates), and the fact that the sum of convex functions is convex. If $O'$ lies inside or on the perimeter of $\triangle ABC$, let ray $\overrightarrow{OP}$ intersect a side of the triangle at $M$, and let $P = \alpha(m)$. Because $g = g_1$ is convex over $[0,m]$, the maximum value of $g$ is attained at either $0$ or $m$. Since $P$ is between $O'$ and $M$, $0 \leq 1 \leq m$, so $f(P) = g(1) \leq \max(g(0), g(m)) = \max(f(O), f(P)) \leq 2R$. If $O'$ lies in the exterior of $\triangle ABC$, let ray $\overrightarrow{OP}$ intersect two sides of the triangle at $Q$ and $R$. Let $Q = \alpha(q)$, $R = \alpha(r)$, and suppose without loss of generality that $q < r$. Because $g = g_1$ is a convex function over $[q,r]$, the maximum value of $g$ is attained at either $q$ or $r$. Since $P$ is between $O'$ and $R$, $q \leq 1 \leq r$, so $f(P) = g(1) \leq \max(g(0), g(r)) = \max(f(O'), f(R)) \leq 2R$, which completes the proof of this lemma. Lemma 4: For any point $P$ in the interior of $\triangle ABC$, $\min(AP, BP, CP) \leq R$. Proof: Let $O'$ be the circumcenter of $\triangle ABC$, and let $O'_A$, $O'_B$, and $O'_C$ be the projections of $O'$ onto $BC$, $CA$, and $AB$, respectively. Without loss of generality, let $P$ lie in the interior of or on the perimeter of $AO'_B O' O'_C$. Then $AP$ must be less than or equal to the length of the diameter of the circumcircle of $AO_B' O' O_C'$. If $O$ is the circumcenter of $\triangle ABC$, then the only if direction of this problem follows immediately from lemma 2. Without loss of generality, let $OC = \max(OA, OB, OC)$. We will show that if $O$ is not the circumcenter, then the perimeter of $\triangle A_1 B_1 C_1$ is less than the perimeter of $\triangle A1B1C$. By lemma 2, this statement is equivalent to $\frac{AO \cdot BC}{2R} + \frac{BO \cdot AC}{2R} < CB_1 + CA_1 = OC(\cos B_1 CO + \cos OCA_1)$. By the law of cosines, $\cos B_1 CO = \cos ACO = \frac{OC^2 + AC^2 - AO^2}{2OC \cdot AC}$ an $\cos A_1 CO = \cos BCO = \frac{OC^2 + BC^2 - OB^2}{2OC \cdot BC}$, so the desired inequality can be rewritten as \[ BC \cdot \frac{AO}{R} + AC \cdot \frac{BO}{R} < \frac{OC^2 - OB^2}{BC} + \frac{OC^2 - OA^2}{AC} + BC + AC.\] By lemma 4, we can suppose without loss of generality that $OA < R$. Then $BC \cdot \frac{AO}{R} < BC$. Furthermore, by the triangle inequality $OC + OB > BC$ and $OC+OA > AC$, so it is enough to show that \[ AC\left(\frac{BO}{R} - 1\right) \leq 2OC - OB - OA. \] If $\frac{BO}{R} - 1 \leq 0$, then we are done since the right-hand side is nonnegative. If $\frac{BO}{R} - 1> 0$, then by lemma 2, $AC\left(\frac{BO}{R} - 1\right) \leq 2BO - 2R$, so it is enough to show \[ 2OC + 2R \geq 3BO + OA. \] But $2OC \geq 2OB$, and by lemma 3, $2R \geq OB + OA$; adding these inequalities yields the desired inequality.

pi37

17.05.2016 10:44

One direction is obvious, so now assume that all three perimeter inequalities hold. Let $X$ be the orthocenter of $A_1B_1C_1$; we will show $O=X$. If this is not the case, $O$ must without loss of generality lie within triangle $B_1OC_1$ (where $O$ can lie on the sides but can't be $X$). This is true since it's easy to show $O$ lies within triangle $A_1B_1C_1$ (if $A_1B_1C_1$ is obtuse, we choose $B_1OC_1$ with $\angle A_1$ obtuse). Now let $x=\angle C_1B_1O$ and let $y=\angle B_1C_1O$. We know $x\le \angle C_1B_1X=90-\angle C_1$ and $y \le \angle B_1C_1X=90-\angle B_1$. This also implies that if $z=180-x-y$, then $z\ge 180-\angle A_1$ (and since $O$ lies in $A_1B_1C_1$, $z\ge \angle A_1$. Now we note the following: \[ \frac{AB_1}{B_1C_1}+\frac{AC_1}{B_1C_1}=\frac{\sin( \angle AC_1B_1 )+\sin(\angle AB_1C_1)}{\sin \angle A} =\frac{\cos(x)+\cos(y)}{\sin z} \]Meanwhile. \[ \frac{A_1B_1}{B_1C_1}+\frac{A_1C_1}{A_1B_1}=\frac{\cos(90-\angle B_1)+\cos(90-\angle C_1)}{\sin(\angle A_1)} \]Now $0<x\le 90-\angle C_1<180$ so $\cos(x)\ge \cos(90-\angle C_1)$. Similarly, $\cos(y)\ge \cos(90-\angle B)$. Meanwhile, $z\ge \text{max}(180-\angle A_1,\angle A_1)$, so $\sin(z)\le \sin(A_1)$ in all cases. Putting these together yields \[ AB_1+AC_1\ge A_1B_1+A_1C_1 \]with equality when $x=90-\angle C_1$, $y=90-\angle B_1$, or equivalently when $O=X$.

596066

21.04.2021 23:41

Diagram :-

Attachments:

pokpokben

26.05.2024 17:53

Zhero wrote: Here's a much longer solution... I'm not certain that this is correct.

Let $R$ be the circumradius of $\triangle ABC$. Lemma 1: $B_1C_1 = \frac{OA \cdot BC}{2R}$. Proof: $AC_1OB_1$ is cyclic, and has diameter $AO$, so by the law of sines $\frac{B_1C_1}{\sin A} = OA$, so $B_1C_1 = OA \sin A = \frac{OA \cdot BC}{2R}$. Lemma 2: $2R \geq \max(AB, BC, CA)$. Proof: This follows trivially from the fact that the length of a longest possible chord in a circle is the diameter. Lemma 3: Let $P$ be any point not in the exterior of $\triangle ABC$. Then $2R \geq PA + PB + PC - \max(PA, PB, PC)$. Proof: Let $f_1(P) = PB+PC$, $f_2(P) = PC+PA$, $f_3(P) = PA+PB$, and $f(P) = \min(f_1(P), f_2(P), f_3(P))$. We wish to show that $f(P) \leq 2R$. We will first prove this when $P$ is on the perimeter of $\triangle ABC$. Without loss of generality, let it be on side $AB$. Then $f(P) = PA + PB + PC - \max(PA,PB,PC) \leq PA + PB = AB \leq 2R$ Suppose now that $O'$ is the circumcenter of $\triangle ABC$. If $P = O'$, then we are immediately done. Otherwise, without loss of generality suppose that $O'$ is the origin, let $u$ be the vector from $O'$ to $P$, and for any nonnegative real $t$ let $\alpha(t) = tu$. (The image of $\alpha$ would be the ray $\overrightarrow{O'P}$.) Let $g_i(t) = f_i(\alpha(t))$ for $i = 1,2,3$ and $g(t) = \min(g_1(t), g_2(t), g_3(t))$. We may suppose without loss of generality that $g_1(1) = g(1)$. We claim that $g_1(t) = g(t)$ for all nonnegative $t$. If $g_1(t) \neq g(t)$ for some $t$, then $g_j(t) < g_1(t)$ for some $j$, which we may suppose without loss of generality equals 2. Because $g_2(1) \geq g_1(1)$, the intermediate value theorem implies the existence of some $c \in [\min(t, 1), \max(t, 1)]$ such that $g_1(c) = g_2(c)$. Noting that $t \neq 0$ (since $g_1(0) = g_2(0)$ while $g_1(t) \neq g_2(t)$), we find that $c \geq \min(t, 1) > 0$. Letting $X = \alpha(c)$, we find that $BX + CX = f_1(X) = f_2(X) = AX + CX$, so $AX = BX$. It follows that $X$ must be equidistant from $A$ and $B$, so $X$ must lie on the perpendicular bisector of $AB$. Because $X$ also lies on the line through $\alpha(t)$ and $O$, we find that either $X = O$ or $\alpha(t)$ lies on the perpendicular bisector of $AB$. The former case is impossible since $c \neq 0$. In the latter case, we find that every point on the image of $\alpha$ is equidistant from $A$ and $B$, so $g_1(t) = g_2(t) < g_1(t)$, which is impossible. Hence, it is safe for us to assume that $g(t) = g_1(t)$ for all nonnegative $t$. We now observe that $g = g_1$ is convex for all nonnegative $t$; this can easily be verified from the fact that as $P$ varies along $\overrightarrow{O'P}$, the distances $PB$ and $PC$ are convex in the $x$-coordinate of $P$ (which can easily be verified with coordinates), and the fact that the sum of convex functions is convex. If $O'$ lies inside or on the perimeter of $\triangle ABC$, let ray $\overrightarrow{OP}$ intersect a side of the triangle at $M$, and let $P = \alpha(m)$. Because $g = g_1$ is convex over $[0,m]$, the maximum value of $g$ is attained at either $0$ or $m$. Since $P$ is between $O'$ and $M$, $0 \leq 1 \leq m$, so $f(P) = g(1) \leq \max(g(0), g(m)) = \max(f(O), f(P)) \leq 2R$. If $O'$ lies in the exterior of $\triangle ABC$, let ray $\overrightarrow{OP}$ intersect two sides of the triangle at $Q$ and $R$. Let $Q = \alpha(q)$, $R = \alpha(r)$, and suppose without loss of generality that $q < r$. Because $g = g_1$ is a convex function over $[q,r]$, the maximum value of $g$ is attained at either $q$ or $r$. Since $P$ is between $O'$ and $R$, $q \leq 1 \leq r$, so $f(P) = g(1) \leq \max(g(0), g(r)) = \max(f(O'), f(R)) \leq 2R$, which completes the proof of this lemma. Lemma 4: For any point $P$ in the interior of $\triangle ABC$, $\min(AP, BP, CP) \leq R$. Proof: Let $O'$ be the circumcenter of $\triangle ABC$, and let $O'_A$, $O'_B$, and $O'_C$ be the projections of $O'$ onto $BC$, $CA$, and $AB$, respectively. Without loss of generality, let $P$ lie in the interior of or on the perimeter of $AO'_B O' O'_C$. Then $AP$ must be less than or equal to the length of the diameter of the circumcircle of $AO_B' O' O_C'$. If $O$ is the circumcenter of $\triangle ABC$, then the only if direction of this problem follows immediately from lemma 2. Without loss of generality, let $OC = \max(OA, OB, OC)$. We will show that if $O$ is not the circumcenter, then the perimeter of $\triangle A_1 B_1 C_1$ is less than the perimeter of $\triangle A1B1C$. By lemma 2, this statement is equivalent to $\frac{AO \cdot BC}{2R} + \frac{BO \cdot AC}{2R} < CB_1 + CA_1 = OC(\cos B_1 CO + \cos OCA_1)$. By the law of cosines, $\cos B_1 CO = \cos ACO = \frac{OC^2 + AC^2 - AO^2}{2OC \cdot AC}$ an $\cos A_1 CO = \cos BCO = \frac{OC^2 + BC^2 - OB^2}{2OC \cdot BC}$, so the desired inequality can be rewritten as \[ BC \cdot \frac{AO}{R} + AC \cdot \frac{BO}{R} < \frac{OC^2 - OB^2}{BC} + \frac{OC^2 - OA^2}{AC} + BC + AC.\] By lemma 4, we can suppose without loss of generality that $OA < R$. Then $BC \cdot \frac{AO}{R} < BC$. Furthermore, by the triangle inequality $OC + OB > BC$ and $OC+OA > AC$, so it is enough to show that \[ AC\left(\frac{BO}{R} - 1\right) \leq 2OC - OB - OA. \]If $\frac{BO}{R} - 1 \leq 0$, then we are done since the right-hand side is nonnegative. If $\frac{BO}{R} - 1> 0$, then by lemma 2, $AC\left(\frac{BO}{R} - 1\right) \leq 2BO - 2R$, so it is enough to show \[ 2OC + 2R \geq 3BO + OA. \]But $2OC \geq 2OB$, and by lemma 3, $2R \geq OB + OA$; adding these inequalities yields the desired inequality.

Actually, the part with $2R>OA+OB$ can be proven easier. Just consider an ellipse with focuses $A$ and $B$ such that $AX+XB=2R$. Obviously, the circumcenter lies on this ellipse, and the point $O$ is captured in a quadrilateral formed by perpendicular bisectors of $CB, AB$; and $AC, AB$. It is easy to conclude that no point inside the quadrilateral lies outside the ellipse, so we are done