Where's the complex bash fam?

I think I can calculate it (maybe in 30 minutes)

First, we add the midpoint of BC, and then we use Ptolemey inequality for KPMQ.

Let $M$ be the midpoint of $BC$. First note that $\angle OMC = \angle OQC = 90$ and so $OMCQ$ is cyclic. Analogously $OMBP$ is cyclic as well. Further, $\angle OQM = \angle OCM = \angle OBM = \angle OPM$ so $MP = MQ$. By Ptolemy's inequality (which famously turns into an equality when it's cyclic), we have $$KP \cdot QM + KQ \cdot PM \geqslant KM \cdot PQ$$this means $KP + KQ \geqslant \frac{KM \cdot PQ}{PM}$. But note that $\triangle MPQ \sim \triangle OBC$, therefore $\frac{PQ}{PM} = \frac{BC}{OB}$ and $KM = AO = OB$ by the parallelogram. Therefore, $KP + KQ \geqslant BC$, as desired. $\blacksquare$

This problem is most likely the reason why they were trying to hide day2 problems for so long.

starchan wrote: Where's the complex bash fam? Let $\triangle ABC$ be inscribed in the unit circle, and let $\ell$ meet the unit circle at $X$ and $Y$, so that $$|a|=|b|=|c|=|x|=1$$$$o=0$$$$h=a+b+c$$$$y=-x$$$$k=\frac{a+h}2 = \frac{2a+b+c}2$$$$p=\frac{x+y+b-xy\overline{b}}2 = \frac{b^2+x^2}{2b}$$$$q=\frac{x+y+c-xy\overline{c}}2 = \frac{c^2+x^2}{2c}$$$$|k-p| = \frac{|2ab+bc-x^2|}2$$$$|k-q| = \frac{|2ac+bc-x^2|}2$$Then letting $Z$ be the point with coordinate $$z = \frac{x^2-bc}{2a}$$we have $|k-p| = |b-z|$ and $|k-q| = |c-z|$. So by the triangle inequality, $|k-p|+|k-q| \ge |b-c|$. $\blacksquare$

nice probl3m but little bashy

Is this correct? Rephrase in terms of reference triangle $M_AM_BM_C$ (midpoints of $3$ sides.) Then the problem becomes: new problem wrote: Let $ABC$ be a triangle and let $A'$ be the antipode of $A$ on $(ABC)$. A line passing through the orthocenter $H$ intersects $(ABH)$ and $(ACH)$ again at $P$ and $Q$. Prove that $A'P + A'Q \geq 2BC$. However, $PQ$ is the Steiner line of some point $R$ on $(ABC)$. Reflect $A'$ over $B, C$ to get $A_1, A_2$. Then since $PA' = RA_1$, $QA' = RA_2$, it suffices to prove $RA_1 + RA_2 \geq 2BC$. But $A_1A_2 = 2BC$ so we are done.

Trig bash anyone? Let $R$ be the radius of $(ABC)$, $M$ the midpoint of $BC$, and $x = MP$. Observe that $(MOPB)$ and $(MOCQ)$ are cyclic, so \[\angle OPM = \angle OBC = \angle OCM = \angle OQM = 90-\angle A\]which means $MQ = x$. Furthermore, since $K, M$ are diametrically opposite on the nine-point circle, $MK = R$. Thus, if $\angle KMP = \theta$, then \[KP + KQ = \sqrt{R^2 + x^2 - 2Rx\cos\theta} + \sqrt{R^2 + x^2 - 2Rx\cos(2A-\theta)}\]which we want to show is at least $BC = 2R\sin A$. $KP + KQ$ achieves minimality when $K = P$, $K = Q$, or the derivative of $KP + KQ$ wrt $\theta$ is $0$. If $K = P$, then \[KP + KQ = \sqrt{R^2 + x^2 - 2Rx\cos 2A} + R-x \geq 2R\sin 2A \Leftrightarrow R(2\sin A - 1) + x \leq \sqrt{R^2 + x^2 - 2Rx\cos 2A}.\]Squaring both sides and simplifying, we get $R(4\sin^2 A - 4\sin A) \leq 2x(2\sin^2 A - 2\sin A)$, which is true since $x \leq R$. We get the same result for $K = Q$. Now, for the last case, \[\frac{d}{d\theta}KP + KQ = 0 \Leftrightarrow \frac{\sin\theta}{\sqrt{R^2 + x^2 - 2RX\cos \theta}} = \frac{\sin(2A-\theta)}{\sqrt{R^2 + x^2 - 2Rx\cos (2A-\theta)}}.\]Let $MK \cap PQ = G$. Since $MP = MQ$, this means $\frac{\sin\theta}{\sin (2A-\theta)} = \frac{PG}{GQ} = \frac{KP}{KQ}$, so $\angle PKM = \angle MKQ$. By Law of sines on $\triangle MPK$ and $\triangle MQK$, we have $\angle KMP = \angle KMQ = \angle A$. Thus, \[KP + KQ \geq MK\sin A + MK\sin A = 2R\sin A.\]Because for all three cases where $KP + KQ$ is minimal still results in $KP + KQ \geq BC$, we conclude $KP + KQ \geq BC$ for all $P, Q,$ and $K$.

Let $M$ be the midpoint of $BC$. Note that $P\in (OB)$ and $Q\in (OC)$, so $\triangle MPQ\sim \triangle OBC$ by spiral similarity and symmetry. Moreover, we have $MK = OA = OB = OC$ from parallelogram $AKMO$. Now, rotate $\triangle MKP$ about $M$ so that $P\to Q$ and $K\to K'$. Then $\triangle MKK'\cong\triangle OBC$, so we have $$KP + KQ = K'Q + KQ \ge KK' = BC$$ as desired.

My method to solve this problem isn't the most elegant (certainly when compared to CT17's tricky solution), but it is very "structural". (EDIT: This isn't quite my original solution, and I only found the Ptolemy Inequality fix by reading through this method before posting my own. So it's hard to say whether I would have actually been able to come up with this solution myself. But oh well.) Instead of fixing $\triangle ABC$ and letting the line $PQ$ vary, we'll fix the line $PQ$ and segment $\overline{BC}$, and instead let point $A$ vary. Let $M$ be the midpoint of $\overline{BC}$; then $K$ moves along a arc with center $M$ and radius $R$, the circumradius of $\triangle ABC$. Also, because $BP$ and $CQ$ are perpendicular to $\ell$, the reflections $B'$ and $C'$ over $\ell$ also lie on $\odot(ABC)$, and in particular quadrilateral $B'BCC'$ is an isosceles trapezoid. This allows us to remove points $H$ and $A$ entirely, and instead consider $K$ to be an arbitrary point for which $MK = R$. [asy][asy] size(200); defaultpen(linewidth(0.6)+fontsize(11)); int r = 60, s = 340; pair O = origin, B = dir(r), C = dir(s), Bp = dir(180-r), Cp = dir(180-s), M = (B+C)/2, P = (B+Bp)/2, Q = (C+Cp)/2; pair K = M + dir(200); draw(arc(M,1,110,250),gray(0.5)+linetype("4 4")); draw(B--C--Cp--Bp--cycle,rgb(0.1,0.1,0.8)); draw(unitcircle, rgb(0.1,0.6,0.1)); draw(P--K--Q,rgb(0.9,0.5,0.3)); draw(M--K,rgb(0.8,0.1,0.1)); dot("$B$",B,NE); dot("$C$",C,SE); dot("$B'$",Bp,NW); dot("$C'$",Cp,SW); dot("$M$",M,NE); dot("$P$",P,SE); dot("$Q$",Q,NE); dot("$K$",K,SW); [/asy][/asy] By Ptolemy's Inequality, \[ MP(KP + KQ) \geq PQ\cdot MK, \]so \[ KP + KQ \geq PQ\cdot \frac{MK}{MP} = PQ\cdot\frac{AC}{2R}. \]All that remains is to show that this last quantity is greater than or equal to $BC$. This allows us to remove points $M$ and $K$ from the picture. It also allows us to discard points $P$ and $Q$, since $PQ$ is just the height of the trapezoid $BCC'B'$. [asy][asy] size(200); defaultpen(linewidth(0.6)+fontsize(11)); int r = 60, s = 340; pair O = origin, B = dir(r), C = dir(s), Bp = dir(180-r), Cp = dir(180-s); pair X = foot(B,C,Cp), Y = -C; draw(B--C--Cp--Bp--cycle,rgb(0.1,0.1,0.8)); draw(unitcircle, rgb(0.1,0.6,0.1)); draw(B--X^^rightanglemark(B,X,C,2),rgb(0.8,0.1,0.1)); draw(C--Y--Bp--C^^rightanglemark(Y,Bp,C,2),rgb(0.9,0.5,0.3)); dot("$B$",B,NE); dot("$C$",C,SE); dot("$B'$",Bp,NW); dot("$C'$",Cp,SW); dot("$X$",X,S); dot("$Y$",Y,NW); [/asy][/asy] Finally, let $X$ be the foot of the perpendicular from $B$ to $\overline{CC'}$, and let $Y$ be the antipode of $C$ with respect to $\odot(BCC'B')$. Triangles $YB'C$ and $CXB$ are similar, so $\tfrac{AC}{2R} = \tfrac{BC}{PQ}$. Multiplying both sides by $PQ$ completes the proof. $\blacksquare$

Let $PQ$ be the real axis, and let the complex numbers $a,b,c$ lie on the unit circle. Then $$p =\frac{b+\frac{1}{b}}{2} , q=\frac{c+\frac{1}{c}}{2}, k=a+\frac{b+c}{2}$$ The inequality is equivalent to $$\left\vert a+\frac{b+c}{2} - \frac{b+\frac{1}{b}}{2} \right\vert +\left\vert a+\frac{b+c}{2} - \frac{c+\frac{1}{c}}{2} \right\vert \geq \left\vert b-c \right\vert$$ $$\iff \left\vert a+\frac{c-\frac{1}{b}}{2} \right\vert + \left\vert a+\frac{b- \frac{1}{c}}{2} \right\vert \geq \left\vert b-c \right\vert$$ $$\iff \vert c \vert \cdot \left\vert a+\frac{b-\frac{1}{c}}{2} \right\vert + \vert b \vert \cdot \left\vert a+\frac{c-\frac{1}{b}}{2} \right\vert \geq \vert a \vert \cdot \left\vert b-c \right\vert$$ $$\iff \left\vert ca+\frac{bc-1}{2} \right\vert + \left\vert ba+\frac{bc-1}{2 } \right\vert \geq \vert ab-ac \vert $$ which is true by triangle inequality.

djmathman wrote: My method to solve this problem isn't the most elegant (certainly when compared to CT17's tricky solution), but it is very "structural". Instead of fixing $\triangle ABC$ and letting the line $PQ$ vary, we'll fix the line $PQ$ and segment $\overline{BC}$, and instead let point $A$ vary. Let $M$ be the midpoint of $\overline{BC}$; then $K$ moves along a arc with center $M$ and radius $R$, the circumradius of $\triangle ABC$. Also, because $BP$ and $CQ$ are perpendicular to $\ell$, the reflections $B'$ and $C'$ over $\ell$ also lie on $\odot(ABC)$, and in particular quadrilateral $B'BCC'$ is an isosceles trapezoid. This allows us to remove points $H$ and $A$ entirely, and instead consider $K$ to be an arbitrary point for which $MK = R$. [asy][asy] size(200); defaultpen(linewidth(0.6)+fontsize(11)); int r = 60, s = 340; pair O = origin, B = dir(r), C = dir(s), Bp = dir(180-r), Cp = dir(180-s), M = (B+C)/2, P = (B+Bp)/2, Q = (C+Cp)/2; pair K = M + dir(200); draw(arc(M,1,110,250),gray(0.5)+linetype("4 4")); draw(B--C--Cp--Bp--cycle,rgb(0.1,0.1,0.8)); draw(unitcircle, rgb(0.1,0.6,0.1)); draw(P--K--Q,rgb(0.9,0.5,0.3)); draw(M--K,rgb(0.8,0.1,0.1)); dot("$B$",B,NE); dot("$C$",C,SE); dot("$B'$",Bp,NW); dot("$C'$",Cp,SW); dot("$M$",M,NE); dot("$P$",P,SE); dot("$Q$",Q,NE); dot("$K$",K,SW); [/asy][/asy] For the next step, it helps to have some motivation. This length $MK$ is strictly greater than $MP = MQ$. If $K$ were allowed to vary on a circle with this smaller radius, the strongest inequality we could get is $KP + KQ \geq PQ$ (achieved with $K = P$ or $K = Q$). This motivates seeing how far this bound can be strengthened with the extra leg room. Let $K_0$ be the point on $\overline{MK}$ such that $MK_0 = MP = MQ$. Then $K_0P + K_0Q \geq PQ$ by the Triangle Inequality, so \begin{align*} KP + KQ &\geq K_0P + K_0Q + 2KK_0 \geq PQ + 2KK_0 \\ &= PQ + 2(R - MP) = PQ + 2R - B'C. \end{align*}All that remains is to show that this last quantity is greater than or equal to $BC$. This allows us to remove points $M$ and $K$ from the picture. It also allows us to discard points $P$ and $Q$, since $PQ$ is just the height of the trapezoid $BCC'B'$. [asy][asy] size(200); defaultpen(linewidth(0.6)+fontsize(11)); int r = 60, s = 340; pair O = origin, B = dir(r), C = dir(s), Bp = dir(180-r), Cp = dir(180-s); pair X = foot(B,C,Cp), Y = -C; draw(B--C--Cp--Bp--cycle,rgb(0.1,0.1,0.8)); draw(unitcircle, rgb(0.1,0.6,0.1)); draw(B--X^^rightanglemark(B,X,C,2),rgb(0.8,0.1,0.1)); draw(C--Y--Bp--C^^rightanglemark(Y,Bp,C,2),rgb(0.9,0.5,0.3)); dot("$B$",B,NE); dot("$C$",C,SE); dot("$B'$",Bp,NW); dot("$C'$",Cp,SW); dot("$X$",X,S); dot("$Y$",Y,NW); [/asy][/asy] Finally, let $X$ be the foot of the perpendicular from $B$ to $\overline{CC'}$, and let $Y$ be the antipode of $C$ with respect to $\odot(BCC'B')$. Triangles $YB'C$ and $CXB$ are similar, so \[ 2R - B'C = BY - B'C \geq BC - BX. \]Adding $BX = PQ$ to both sides yields the desired inequality, completing the proof. $\blacksquare$ Isn't $KP+KQ\le K_0P+K_0Q+2KK_0$?

removed by user

hakN wrote:

That's a good solution.

gvole wrote: Isn't $KP+KQ\le K_0P+K_0Q+2KK_0$? Oops, yea, you're right My proof still works, but I have to replace the Triangle Inequality with Ptolemy's Inequality, which makes me sad. Fixed, thanks.

Does my proof work? Note that $\ell$ is perpendicular to the line $OH,$ which is parallel to $BC.$ Therefore, $PQ$ is parallel to $BC.$ Let $M$ be the midpoint of $BC.$ Since $\angle AHB = 180^\circ - \angle ACB,$ quadrilateral $ABCH$ is cyclic. It follows that $OK=OM.$ Also, note that $$PK = PH - HK = 2 \cdot \frac{AH}{2} - \frac{AH}{2} = \frac{AH}{2}$$and similarly $QK = \frac{AH}{2}.$ Therefore, we have $$KP+KQ = AH.$$Since $\angle AQH = 90^\circ = \angle APH,$ we have $APQH$ is cyclic. Thus, $\angle AQP = \angle AHP = \angle ABC$ and $\angle APQ = \angle AHQ = \angle ACB.$ It follows that $$\angle QPM = \angle AQP - \angle AQB = \angle ABC - \angle ACB = \angle BAC.$$Therefore, triangle $QPM$ is similar to triangle $ABC$ with ratio $\frac{1}{2}.$ This implies $PM = \frac{BC}{2}.$ Now, we have $$KP+KQ = AH \ge 2OM = BC.$$The last inequality follows from $OH \le 3R$ (Euler's inequality), where $R$ is the circumradius of $ABC.$ We are done.

We use complex numbers. WLOG let $\ell$ be the real axis and $(ABC)$ be the unit circle, so $o=0$, $h=a+b+c$, $k=a+\frac{b+c}{2}$, $p=\frac{b+\tfrac{1}{b}}{2}$, and $q=\frac{c+\tfrac{1}{c}}{2}$. Then we can calculate \begin{align*} |k-p|&=\left|a+\frac{c-\frac{1}{b}}{2}\right|=\left|b+\frac{\frac{cb}{a}-\frac{1}{a}}{2}\right|,\\ |k-q|&=\left|a+\frac{b-\frac{1}{c}}{2}\right|=\left|c+\frac{\frac{cb}{a}-\frac{1}{a}}{2}\right|,\\ \end{align*}where we multiply by $b/a$ and $c/a$ respectively, both of which have absolute value $1$. The conclusion then follows by triangle inequality on $b$, $c$, and $-\frac{\frac{cb}{a}-\frac{1}{a}}{2}$. $\blacksquare$

sqing wrote: An acute triangle $ABC$ is given and $H$ and $O$ be its orthocenter and circumcenter respectively. Let $K$ be the midpoint of $AH$ and $\ell$ be a line through $O. $ Let $P$ and $Q$ be the projections of $B$ and $C$ on $\ell. $ Prove that$$KP+KQ\ge BC$$ A very good problem . (Russia, Vasily Mokin)

Solved with eisirrational, goodbear, nukelauncher, and Th3Numb3rThr33. Let \((ABC)\) be the unit circle and \(\ell\) the real axis. Then \(K=\frac12(2a+b+c)\), \(p=\frac12(b+1/b)\), and \(q=\frac12(c+1/c)\), so \begin{align*} KP+KQ&=\left\lvert a-\frac1{2b}+c\right\rvert +\left\lvert a+b-\frac1{2c}\right\rvert\\ &=\left\lvert b-\frac1{2a}+\frac{bc}a\right\rvert +\left\lvert c+\frac{bc}a-\frac1{2a}\right\rvert\\ &\ge|b-c|=BC, \end{align*}as desired.

Let $M$ be the midpoint of $|BC|$. Since $OPBM$ and $OQCM$ are cyclic, we have $$\angle{OPM}=\angle{OBM}=\angle{OCM}=\angle{OQM}$$and so $\triangle{PMQ}$ is isosceles and is similar to $\triangle{BOC}$. Note that $|KM|=|OB|=|OC|$, so by Ptolemy's inequality we must have $$KP+KQ\geq \frac{PQ}{MQ}\times OB=\frac{BC}{OB}\times OB=BC.$$Done.

Solution ($100$th Post): Let $X$ be the midpoint of $BC$ . Now let's proceed with $3$ claims. Claim 1: $XBPO$ is a cyclic quadrilateral. Proof : Observe that $$\angle{BXO}=\frac{\pi}{2}=\angle{BPO}$$so $XBPO$ is a cyclic quadrilateral. Claim 2:$XCQO$ is cyclic quadrilateral. Proof : Observe that at $$\angle{CXO} = \angle{OQC} = \frac{\pi}{2}$$so $XCQO$ is cyclic. Claim 3: $\triangle{QXP} \sim \triangle{BOC}$ Proof : Angle Chase from Claim $1$ and Claim $2$ yields $$\angle{OQX}=\angle{OCX}=\angle{XBO}=\angle{XPO}$$so Claim $3$ is proved. Hence $\frac{PQ}{PM} = \frac{BC}{OB}$ and $MK=OA=OB$ . Finish : Using Ptolemy's Inequality we get $$KP+KQ \geq \frac{MK.PQ}{MP}=\frac{BC.OB}{OB}=BC$$So we are done !

We first define a lot of points: Let $M$ be the midpoint of $BC$. Let $N$ be the nine-point center of $\triangle ABC$. Let $E$ and $F$ be the foot from $B$ and $C$ to $AC$ and $AB$, respectively. The first observation is that $\square BPOM$ and $\square BQOM$ are concyclic. This gives $\angle MPQ = \angle OBC$ and $\angle MQP = \angle OCB$, so $\triangle MPQ$ is isosceles and similar to $\triangle BOC$. Now, let $P'$ and $Q'$ be the reflections of $P$ and $Q$ across $N$. Observe that $$\triangle KP'Q'\cong\triangle MPQ\sim\triangle OBC\sim\triangle KFE\implies \triangle KFP'\cong\triangle KQ'E,$$and we have to show that $MP'+MQ'\geq BC$. We are almost done. Let $M'$ be the reflection of $M$ across $E$. Observe that $\triangle KFP'\cup M\cong\triangle KEQ'\cup M'$, so $MP'=M'Q'$. Hence, we have $$MP' + MQ' = M'Q' + MQ' \geq MM' = 2ME = BC,$$done.

Let $\ell$ be the real axis and $(ABC)$ be the unit circle. Then $k = a+\frac b2+\frac c2$ and $p = \frac{b-\frac 1b}2$, $q = \frac{c-\frac 1c}2$. Suffices $$KP+KQ = \left|a+\frac c2+\frac 1{2b}\right|+\left|a+\frac b2+\frac 1{2c}\right| = \left|ab+\frac{bc}2+\frac 12\right| + \left|ac+\frac{bc}2 + \frac 12\right| \geq |ab-ac| = |b-c| = BC$$by triangle inequality.

i solved it by establishing a coordinate system XD thats not so difficult to calculate Excuse me for the comment like this but that's so difficult to type it here(using LaTeX), especially with a mobile

Let $(ABC)$ be the unit circle and let $l$ be the real axis in our plane. Hence, $|a|=|b|=|c|=1$, $q=\overline{q}$ and $p=\overline{p}.$ Note that $k=\frac{h+a}{2}=\frac{2a+b+c}{2}.$ From $BP\perp PO$, we obtain that $$\frac{b-p}{p}=-\overline{\left(\frac{b-p}{p}\right)}\iff \overline{b}-p=-b+p \iff p=\frac{b+\frac{1}{b}}{2}.$$Similarly, we get that $$q=\frac{c+\frac{1}{c}}{2}.$$Now we have that $$KP=|k-p|=\left\lvert a+\frac{b}{2}+\frac{c}{2}-\frac{b}{2}-\frac{1}{2b}\right\rvert \cdot \left\lvert\frac{b}{a}\right\rvert=\left\lvert b+\frac{bc}{2a}-\frac{1}{2a}\right\rvert,$$$$KQ=|k-q|=\left\lvert a+\frac{b}{2}+\frac{c}{2}-\frac{c}{2}-\frac{1}{2c}\right\rvert \cdot \left\lvert\frac{c}{a}\right\rvert=\left\lvert c+\frac{bc}{2a}-\frac{1}{2a}\right\rvert.$$ Hence, $$KP+KQ=|k-p|+|k-q|=\left\lvert b+\frac{bc}{2a}-\frac{1}{2a}\right\rvert+\left\lvert c+\frac{bc}{2a}-\frac{1}{2a}\right\rvert\ge |b-c|=BC,$$so we are done.

Let $M$ be the midpoint of $BC$. First note $\angle MPO=\angle MBO=\angle OCM=\angle OQM$ implies $MP=MQ$. Additionally we get $\triangle MPQ\sim\triangle OCB$. Fix some choice of $P,Q$ and let $K$ vary on the circle $\Omega$ centered at $M$ with radius equal to the circumradius. Then if $KP+KQ$ is an extreme, then the ellipse with foci $P,Q$ through $K$ is tangent to $\Omega$. By the reflection property, $\angle PKM=\angle MKQ$. Now by SSA on $\triangle PKM,\triangle QKM$ we get $\angle KPM=\angle MQK$ or $\angle KPM=180^\circ-\angle MQK$, so we have two cases. In the first case, $\triangle PKM\cong\triangle QKM$. WLOG $PQ$ intersects $BC$ to the right of $M$. Then if $MK$ intersects $(MOC)$ at $X$ we have $BP\parallel CQ\parallel MX\perp PQ$, and $\angle PXM=\angle MXQ=\angle CMX$, so $PXCM$ is a parallelogram. Then we have $KP+KQ=2KP\ge 2XP=2CM=BC$ as desired. In the second case, $K,P,M,Q$ are cyclic. Then Ptolemy's on $KPMQ$ implies $KP+KQ=KM\cdot\frac{PQ}{PM}=OC\cdot\frac{BC}{OC}=BC$ as desired.