In an acute triangle $ABC$, let $D$, $E$, $F$ be the feet of the perpendiculars from the points $A$, $B$, $C$ to the lines $BC$, $CA$, $AB$, respectively, and let $P$, $Q$, $R$ be the feet of the perpendiculars from the points $A$, $B$, $C$ to the lines $EF$, $FD$, $DE$, respectively. Prove that $p\left(ABC\right)p\left(PQR\right) \ge \left(p\left(DEF\right)\right)^{2}$, where $p\left(T\right)$ denotes the perimeter of triangle $T$ . Proposed by Hojoo Lee, Korea
Problem
Source: IMO Shortlist 2005, problem G7, created by Hojoo Lee
Tags: geometry, inequalities, circumcircle, IMO Shortlist
treegoner
17.08.2006 07:07
Notice that $p(ABC) = 8R \prod cos{\frac{A}{2}}$, $p(DEF) = 8R \prod sinA$, $p(PQR) = 4R \prod sinA$.
Virgil Nicula
20.08.2006 04:10
treegoner wrote: Notice that ........ $\boxed{\ p(PQR) = 4R \prod sinA\ }$. Why ? I found $\boxed{\ p(PQR)=R\sum \sin 2A\sqrt{1-\sin 2B\sin 2C}\ }\ \ !$.
treegoner
20.08.2006 08:39
Sorry! I misread the problem . I will spend some time for this problem later.
Sung-yoon Kim
20.08.2006 09:43
Amir.S wrote: In an acute triangle $ABC$, let $D, E, F, P, Q, R$ be the feet of perpendiculars from $A, B, C, A, B, C$ to $BC, CA, AB, EF, FD, DE,$ respectively. Prove that $p(ABC)p(PQR) \ge p(DEF)^{2}$, where $p(T )$ denotes the perimeter of triangle $T$ . $BC=a$, $QR= \cos A \sqrt{a^{2}-4bc \cos B \cos C \sin^{2}A}$, $EF=a \cos A$, so $BC \cdot QR \geq EF^{2}\iff a \cos A \sqrt{a^{2}-4bc \cos B \cos C \sin^{2}A}\geq a^{2}\cos^{2}A$ $\iff \sqrt{a^{2}-4bc \cos B \cos C \sin^{2}A}\geq a \cos A \iff a^{2}\sin^{2}A \geq 4bc \cos B \cos C \sin^{2}A$ $\iff a^{2}\geq 4bc \cos B \cos C$, which is true as $a=b \cos C+c \cos B$. So $p(ABC)p(PQR)=(BC+CA+AB)(QR+RP+PQ)$ $\geq (\sqrt{BC \cdot QR}+\sqrt{CA \cdot RP}+\sqrt{AB \cdot PQ})^{2}\geq (EF+FD+DE)^{2}=p(DEF)^{2}$.
ideahitme
20.08.2006 10:16
This is G7 in ISL 2005. It was proposed by me The key idea is to use the inequality Sung-yoon stated. Check out the proof in the new edition of my weblication 'Topics in Inequalities - Theorems and Techniques' which is now availabe at my new homepage at http://ultrametric.googlepages.com/tin2006new.pdf In fact, more is true. Look at the solutions in the official IMO short list book.
The QuattoMaster 6000
14.04.2009 04:13
Let $ \angle BAC = A$, $ \angle ABC = B$, $ \angle ACB = C$, $ BC = a$, $ AC = b$, and $ AB = c$. Furthermore, let $ O$ be the circumcenter of $ \triangle ABC$. Since $ \angle OAC = 90 - \angle ABC = 90 - \angle AEF = \angle PAE$, we see that $ APO$ is a line. Similarly, $ BQO$ and $ CRO$ are lines. Now, let the midpoints of $ AC$, $ BC$, and $ AB$ be $ Y$, $ X$, and $ Z$ respectively. Notice that since $ OZ\perp BC$, we have that $ OD \ge OZ$. Since $ OR\perp DE$ and $ OQ\perp DF$, we have that $ OQDR$ is cyclic with diameter $ OD$, so\[ QR = OD\sin \angle QOR \ge OZ\sin 2 A = 2\sin A\cos A\cdot \frac {BZ\cos A}{\sin A} = a\cos^2 A \]Doing the same with the others, we get that\[ (QR + PQ + PR)(BC + AB + AC) \ge (\sum a\cos^2 A)(\sum a)\ge (\sum a\cos A)^2\]by Cauchy. Notice that $ \triangle AFE \sim \triangle ACB$ with ratio $ \frac {AE}{AB} = \cos A$. Hence, $ \frac {FE}{BC} = \cos A\implies FE = a\cos A$ and we may conclude that $ p(ABC)p(PQR)\ge [p(DEF)]^2$.
Zhero
25.02.2011 03:40
In fact, a much stronger inequality is true. According to Fagnano's problem, the perimeter of $\triangle DEF$ is less than or equal to the perimeter of the medial triangle, which has perimeter $\frac{p(ABC)}{2}$. It is therefore sufficient to show that $p(DEF) \leq 2 p(PQR)$.
Because $\triangle DEF$ is the orthic triangle of $\triangle ABC$, $C$ is the $F$-excenter of $\triangle DEF$, so $P$ is the point where the $F$-excircle is tangent to $\triangle FED$. Similarly, $Q$ and $R$ are the tangency points of the $E$-excircle and $D$-excircle to $FD$ and $EF$, respectively. Hence, we may let $PE = QF = x$, $RF = PD = y$, and $QD = RE = z$. We wish to show that $x+y+z \leq PQ + QR + RP$.
By the law of cosines,
\begin{align*}
\cos D
&= \frac{(x+y)^2 + (z+x)^2 - (y+z)^2}{2(x+y)(x+z)} \\
&= \frac{2x^2 + 2xy + 2xz- 2yz}{2(x+y)(x+z)} \\
&= 1 - \frac{2yz}{(x+y)(x+z)},
\end{align*}
so $PQ = \sqrt{y^2 + z^2 - 2 \left(1 - \frac{2yz}{(x+y)(x+z)}\right)} = \sqrt{(y-z)^2 + \frac{4y^2 z^2}{(x+y)(x+z)}}$. $PQ + QR + RP \geq x+y+z$ therefore becomes
\[ \sum_{cyc} \sqrt{(y-z)^2 + \frac{4y^2 z^2}{(x+y)(x+z)}} \geq x+y+z, \]
or
\[ \sum_{cyc} \sqrt{(y-z)^2(x+y)(y+z)(z+x) + 4y^2 z^2 (y+z)} \geq \sqrt{(x+y)(y+z)(z+x)}(x+y+z).\]
Squaring both sides, our inequality becomes
\[\sum_{cyc} (y-z)^2(x+y)(y+z)(z+x) + \sum_{cyc} 4y^2 z^2 (y+z) + 2A \geq (x+y)(y+z)(z+x)(x+y+z)^2,\]
where
\begin{align*}
A
&= \sum_{cyc} \sqrt{(y-z)^2 (x+y)(y+z)(z+x) + 4y^2 z^2 (y+z)} \sqrt{(x-y)^2 (x+y)(y+z)(z+x) + 4x^2 y^2 (x+y)} \\
&\geq \sum_{cyc} \sqrt{4y^2 z^2 (y+z)} \sqrt{4x^2 y^2 (x+y)} \\
&= 4xyz \sum_{cyc} y \sqrt{(y+x)(y+z)}.
\end{align*}
Observing that $\sqrt{(a+b)(a+c)} \geq a + \sqrt{bc} \iff a(\sqrt{b} - \sqrt{c})^2 \geq 0$, we find that
\[ 4xyz \sum_{cyc} y \sqrt{(y+x)(y+z)} \geq \sum_{cyc} 4x^3 yz + \sum_{cyc} 4x^{\frac{3}{2}} y^{\frac{3}{2}} z^2, \]
so it is enough to show that
\[ \sum_{cyc} (y-z)^2(x+y)(y+z)(z+x) + \sum_{cyc} 4y^2 z^2 (y+z) + 8 \sum_{cyc} x^3 yz + 8 \sum_{cyc} x^{\frac{3}{2}} y^{\frac{3}{2}} z^2 \geq (x+y)(y+z)(z+x)(x+y+z)^2. \]
We also have
\begin{align*}
(x+y)(y+z)(z+x) \left( \left( \sum_{cyc} (y-z)^2\right) -(x+y+z)^2 \right)
&= (x+y)(y+z)(z+x)(x^2 + y^2 + z^2 - 4xy - 4yz - 4xz) \\
&= (\sum_{sym} x^2 y + 2 xyz)(x^2 + y^2 + z^2 - 4(xy + yz + zx)) \\
&= \sum_{sym} x^4 y + \sum_{sym} x^3 y^2 + \sum_{sym} x^2 y^2 z + \sum_{sym}x^3 yz - 4 \sum_{sym} x^2 y^2 z \\
&\, \, \, \, -4\sum_{sym}x^3 y^2 - 4 \sum_{sym} x^2 y^2 z - 4 \sum_{sym} x^3 yz \\
&= \sum_{sym} x^4 y - 3 \sum_{sym} x^3 y^2 - 7 \sum_{sym} x^2 y^2 z - 3 \sum_{sym} x^3 yz
\end{align*}
Henceforth, we shall denote by $[a,b,c]$ the sum $\sum_{sym} x^a y^b z^c$. Then we may thus rewrite the desired inequality as
\[ [4,1,0] + 4[3,2,0] + 4[3,1,1] + 4[2,\frac{3}{2}, \frac{3}{2}] \geq 3[3,2,0] + 7[2,2,1] + 3[3,1,1], \]
which is clearly equivalent to
\[ [8,2,0] + [6,4,0] + [6,2,2] + 4[4,3,3] \geq 7[4,4,2]. \]
By Schur's inequality with $r=2$, we have $[4,0,0] + [2,1,1] \geq 2[3,1,0]$, so $[4,3,3] \geq 2[5,3,2] - [6,2,2]$. Hence, it is enough to show that
\[ [8,2,0] + [6,4,0] + [6,2,2] + 8[5,3,2] \geq 4[6,2,2] + 7[4,4,2]. \]
By Muirhead's inequality, $[5,3,2] \geq [4,4,2]$ and $[6,4,0] \geq [6,2,2]$, so it is sufficient to show
\[ [8,2,0] + [5,3,2] \geq 2[6,2,2]. \]
Now, we have
\[ \sum_{cyc} (x^2 - y^2)^2 (x^4 + y^4) z^2 \geq 0,\]
which expands to
\[ [8,2,0] \geq 2[6,2,2] - [4,4,2]. \]
Thus, it is sufficient to show
\[ 2[6,2,2] - [4,4,2] + [5,3,2] \geq 2[6,2,2], \]
or
\[ [5,3,2] \geq [4,4,2], \]
which follows from Muirhead.
(dnkywin informed me that this stronger result is true.)
NewAutumn
25.02.2011 06:56
Edit: clear.
sayantanchakraborty
26.05.2014 07:46
We note that $\angle{PAF}=\angle{QBF}=90-C$ so the lines $AP,BQ,CR$ concur at the circumcenter $O$ of $\triangle{ABC}$.Thus we note that $POQF$ is a cyclic quadrilateral with $OF$ as its diameter,so $PQ=OF\sin2C \ge d(O,AB)\sin2C =R\cosh\sin2C$.We get $PR$ and $QR$ analogously.We also note that $p(ABC)=2R\sum{sinA}$ and $p(DEF)=R\sum{sin2A}$(Note that the pedal triangle has circumradius $\frac{R}{2}$ so applying sine rule will make the fact clear).Thus the problem is equivalent to showing that $2 \sum{sinA} \sum{sin2CcosC} \ge (\sum{sin2A})^2$. $\Leftrightarrow (\sum{(\sqrt{2sinA})^2})(\sum{(\sqrt{sin2AcosA})^2}) \ge (\sum{sin2A})^2$ which is obvious by Cauchy.It can easily be seen that equality holds if and only if $\triangle{ABC}$ is equilateral.
AnonymousBunny
04.06.2014 20:20
Surprisingly simple for a G7.
We adopt the standard conventions $BC=a, CA= b, AB=c, \angle BAC = A, \angle ABC = B, \angle BCA = C.$ Let $R$ be the circumradius of $\triangle ABC$ and let $O$ be its circumcenter.
Claim: $AP, BQ, CR$ are concurrent at $O.$
Proof: Note that $BCEF$ is cyclic since $CF \perp AB$ and $BE \perp AC.$ Hence, $\angle PBC = 90^{\circ} - \angle BDF = 90^{\circ} - A = \angle OBC,$ so $A,P,O$ are collinear. Similarly $B,Q,O$ and $C,R,O$ are also collinear. The result follows. $\blacksquare$
Since $OP \perp DP$ and $OQ \perp DQ,$ $OPDQ$ is cyclic with diameter $OD.$ By sine rule, $PQ = OD \sin POQ = OD \sin 2A.$ Let $M$ be the midpoint of $BC.$ Since $OM \perp BC,$ $OD \geq OM.$ From right angled $\triangle OMB,$ $OM= R \sin (90^{\circ} - A) = R \cos A.$ Hence, $PQ \geq R \sin 2A \cos A.$ We get similar inequalities for $QR$ and $RP.$
By sine rule on $\triangle DEF,$ $\dfrac{DF}{\sin DEF} = R \implies \dfrac{DF}{\sin (\pi - 2A)} = R \implies DF = R \sin 2A,$ where we have used the fact that the radius of the nine point circle is half the radius of the circumcircle (an alternative way to compute $DF$ would be to use sine rule on $\triangle BDF,$ but that is slightly more tedious). We obtain analogous expressions for $EF$ and $DE.$
The desired inequality transforms to
\[(2 \sin A + 2 \sin B + 2 \sin C)(\sin 2A \cos A + \sin 2B \cos B + \sin 2C \cos C) \geq (\sin 2A + \sin 2B + \sin 2C)^2 \\
\iff \\ (\sin A + \sin B + \sin C)(\sin A \cos ^2 A + \sin B \cos ^2 B + \sin C \cos ^2 C) \geq (\sin A \cos A + \sin B \cos B + \sin C \cos C)^2,\]
which is just Cauchy Schwartz on the sequences $(\sqrt{\sin A}, \sqrt{\sin B}, \sqrt{\sin C})$ and $(\cos A \sqrt{\sin A}, \cos B \sqrt{\sin B}, \cos C \sqrt{\sin C}).$
For equality to hold, the midpoints must coincide with the feet of perpendiculars, so $\triangle ABC$ must be equilateral. Plugging $A=B=C = \dfrac{\pi}{3}$ in this equation, we see that equality indeed does hold. Thus, equality holds iff $\triangle ABC$ is equilateral. $\blacksquare$
Nanas
26.08.2015 01:13
Sorry ... That was mistaken
AMN300
25.12.2015 19:54
It's about time. Lemma 1: In a triangle $ABC$, we have $\sin A + \sin B + \sin C \ge \sin 2A + \sin 2B+ \sin 2C$ Proof: Sum to product yields $\frac{\sin 2A + \sin 2B}{2} = \sin(A+B) \cos (A-B)$. Now just sum up cyclically. Lemma 2: In a triangle $ABC$ with excircles touching sides $BC, CA, AB$ at $D, E, F$, respectively, the perimeter of triangle $ABC$ is at most twice that of triangle $DEF$ (USA TSTST 2011, P7). Proof: Consider the projection of $EF$ onto side $BC$. Suppose $E$ is projected to point $E'$ and $F$ is projected to point $F'$ on side $BC$. Basic triangle facts yield that $EF \ge E'F'=a-(s-a)(\cos B + \cos C)$. So we simply want \[ \sum_{cyc} a_(s-a)(\cos B + \cos C) \ge \frac{a+b+c}{2} \]. This rearranges to \[ a+b+c \ge 2a \cos A + 2b \cos B + 2c \cos C \]By LoS, it remains to show \[ \sin A + \sin B + \sin C \ge \sin 2A + \sin 2B + \sin 2C \]Observe that sum to product yields $\frac{\sin 2A + \sin 2B}{2} = \sin(A+B) \cos (A-B) \le \sin C$, from which summing yields the desired. First, I claim $p(ABC) \ge 2 p(DEF)$. By extended LoS, we have $EF=2R \cos A \sin A$. Thus we want to show that \[ 2R \sin A + 2R \sin B + 2R \sin C \ge 2 (2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C) \]. Cancelling $2R$ and using $2\sin A \cos A = \sin 2A$, the inequality reduces to Lemma 1 . Next, I claim $p(PQR) \ge \frac{1}{2} p(DEF)$. Let's shift our reference frame to triangle $DEF$. Observe that $P, Q, R$ are just the touch points of the $D, E, F$ excenters (which are $A, B, C$ respectively) in triangle $DEF$. Then this inequality is true by Lemma 2 . Now multiplying our two claims, we have $p(ABC) p(PQR) \ge (p(DEF))^2$, as requested.
anantmudgal09
07.03.2017 23:42
Quite straight-forward for a G7 Note that $P, Q, R$ are the projections of the circumcenter $O$ (of $\triangle ABC$) onto the lines $EF, FD, DE$ respectively. Secondly, $$QR=OD\cdot \sin \angle EDF=OD\cdot \sin 2A \ge 2R\sin A\cos^2A=BC\cos^2A.$$Finally, we observe that $\frac{EF}{BC}=\cos A$ so $$BC\cdot QR \ge EF^2$$and the given inequality follows from the CS inequality. Equality holds for equilateral triangles only.
william122
08.08.2020 22:26
Project $Q,R$ onto $BC$ at $Q',R'$. Note that $Q'D=\cos^2 A\cdot BD$, and $R'D=\cos^2 A\cdot CD$, so $QR\ge Q'R'=BC\cos^2 A$. Also, by similar triangles, $EF=BC\cdot \frac{AE}{BE}=BC\cos A$. So, $$QR\cdot BC\ge Q'R'\cdot BC=BC^2\cos^2 A =EF^2$$Now, by Cauchy Schwarz, $$(PQ+QR+RP)(AB+BC+CA)\ge (P'Q'+Q'R'+R'P')(AB+BC+CA)\ge (DE+EF+FD)^2$$as desired.
sqing
21.10.2020 03:09
AMN300 wrote:
It's about time.
Lemma 1: In a triangle $ABC$, we have $\sin A + \sin B + \sin C \ge \sin 2A + \sin 2B+ \sin 2C$
Proof: Sum to product yields $\frac{\sin 2A + \sin 2B}{2} = \sin(A+B) \cos (A-B)$. Now just sum up cyclically.
Lemma 2: In a triangle $ABC$ with excircles touching sides $BC, CA, AB$ at $D, E, F$, respectively, the perimeter of triangle $ABC$ is at most twice that of triangle $DEF$ (USA TSTST 2011, P7).
Proof: Consider the projection of $EF$ onto side $BC$. Suppose $E$ is projected to point $E'$ and $F$ is projected to point $F'$ on side $BC$. Basic triangle facts yield that $EF \ge E'F'=a-(s-a)(\cos B + \cos C)$. So we simply want \[ \sum_{cyc} a_(s-a)(\cos B + \cos C) \ge \frac{a+b+c}{2} \]. This rearranges to
By LoS, it remains to show \[ \sin A + \sin B + \sin C \ge \sin 2A + \sin 2B + \sin 2C \]Observe that sum to product yields $\frac{\sin 2A + \sin 2B}{2} = \sin(A+B) \cos (A-B) \le \sin C$, from which summing yields the desired.
First, I claim $p(ABC) \ge 2 p(DEF)$. By extended LoS, we have $EF=2R \cos A \sin A$. Thus we want to show that \[ 2R \sin A + 2R \sin B + 2R \sin C \ge 2 (2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C) \]. Cancelling $2R$ and using $2\sin A \cos A = \sin 2A$, the inequality reduces to Lemma 1 .
Next, I claim $p(PQR) \ge \frac{1}{2} p(DEF)$. Let's shift our reference frame to triangle $DEF$. Observe that $P, Q, R$ are just the touch points of the $D, E, F$ excenters (which are $A, B, C$ respectively) in triangle $DEF$. Then this inequality is true by Lemma 2 .
Now multiplying our two claims, we have $p(ABC) p(PQR) \ge (p(DEF))^2$, as requested.
$$a \sin\frac{A}{2}+b \sin\frac{B}{2}+c \sin\frac{C}{2}\geq \frac{a+b+c}{2}\geq a \cos A + b \cos B + c \cos C$$$$\sqrt{1+sinBsinC}\cdot sinA+\sqrt{1+sinCsinA}\cdot sinB+\sqrt{1-sinAsinB}\cdot sinC> sinA+ sinB+ sinC$$
Attachments:
Modesti
30.12.2020 20:55
The key idea is the same, but I saw it by spiral similarity Main claim. $BC\cdot QR\ge EF^2$, and the analogous results hold. Proof. Let $M$ be the midpoint of $BC$. [asy][asy] defaultpen(fontsize(10)); size(220); pair A=dir(70); pair C=dir(-30); pair B=dir(210); pair D=foot(A,B,C); pair E=foot(B,A,C); pair F=foot(C,B,A); pair P=foot(A,F,E); pair Q=foot(B,D,F); pair R=foot(C,D,E); pair M=midpoint(B--C); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(E)); dot("$M$", M, dir(M)); draw(A--B--C--cycle); draw(D--F); draw(D--E); draw(B--Q); draw(M--Q, dashed+blue); draw(M--R, dashed+blue); draw(M--F, dashed+blue); draw(M--E, dashed+blue); draw(E--F, blue); draw(R--Q, blue); draw(C--R); draw(circumcircle(D,E,F), yellow); [/asy][/asy] As $M$ is the midpoint of the arc $\widehat{EMF}$ on the nine-point circle and $Q$, $R$ are the points in which the $E$ and $F-$excircles touch $FD$ and $DE$, respectively, we know that $M$ is the center of the spiral similarity that sends $QR$ to $EF$. Thus, $$\dfrac{QR}{EF}=\dfrac{QM}{MF}=\dfrac{BM\cdot \frac{\sin{\angle MBQ}}{\sin{\angle BQM}}}{\frac{BC}{2}}=\dfrac{\cos{A}}{\sin{\angle BQM}}\ge\cos{A}=\dfrac{EF}{BC}$$Hence, $BC\cdot QR\ge EF^2$, as desired. Finally, by Cauchy-Schwarz Inequality and the above claim, $$p(ABC)\cdot p(PQR)=(BC+CA+AB)(QR+RP+PQ)\ge (\sqrt{BC\cdot QR}+\sqrt{CA\cdot RP}+\sqrt{AB\cdot PQ})^2\ge(EF+FD+DE)^2=p(DEF)^2$$and we are done. $\square$