Let $A_{1}A_{2}A_{3}$ be an acute triangle, and let $O$ and $H$ be its circumcenter and orthocenter, respectively. For $1\leq i \leq 3$, points $P_{i}$ and $Q_{i}$ lie on lines $OA_{i}$ and $A_{i+1}A_{i+2}$ (where $A_{i+3}=A_{i}$), respectively, such that $OP_{i}HQ_{i}$ is a parallelogram. Prove that \[\frac{OQ_{1}}{OP_{1}}+\frac{OQ_{2}}{OP_{2}}+\frac{OQ_{3}}{OP_{3}}\geq 3.\]
Problem
Source: USA TST 2005, Problem 2
Tags: geometry, circumcircle, parallelogram, trigonometry, inequalities, function, geometric transformation
gemath
15.05.2007 10:07
Change notation: triangle $ABC$ circumcenter $(O)$ orthocenter $H,A_{1}\in OA,A_{2}\in BC$ parallelogram $OA_{1}HA_{2}$ let line $HA$ cut $BC$ at $A'$ and $A''$ be mid point of $BC$ then $\angle (HA_{2},BC)=90-B+C\Rightarrow HA_{2}=\frac{HA'}{\sin(90-B+C)}$ and $HA'= HB\cos C=2R\cos B\cos C$ therefore $HA_{2}=\frac{2R\cos B\cos C}{\cos(B-C)}$ and $OA_{2}\ge OA''=R\cos A$ thus $\sum_{a,b,c}\frac{OA_{2}}{OA_{1}}=\sum\frac{OA_{2}}{HA_{2}}\ge\sum\frac{R\cos A}{\frac{2R\cos B\cos C}{\cos(B-C)}}=\sum\frac{\cos A\cos(B-C)}{2\cos B\cos C}\ge\frac{3}{2}\sqrt[3]{\frac{\prod\cos(B-C)}{\cos A}}\ge 3$ because we have $\prod\cos(B-C)=\prod\frac{\sin(B+C)\cos(B-C)}{\sin A}=\frac{\prod(\sin 2B+\sin 2C)}{8\prod\sin A}\ge\frac{8\prod\sin 2A}{8\prod\sin A}=8\prod\cos A$ We are done.
epitomy01
11.12.2007 11:46
Here's a slightly different approach: (use GeMath's notation). Considering triangle $ HBA_{2}$, which has angles $ \angle HBA_{2} = 90 - C$, $ \angle HA_{2}B = C - B + 90$, $ \angle BHA_{2} = B$, and using the fact that $ BH = 2R cos B$ (which can be derived from Sin Rule in AHB), gives us, using Sin Rule: $ HA_{2} = \frac {2R cos B cos C}{cos (B - C)}$ $ BA_{2} = \frac {R sin 2B} {cos (B - C)}$ Using Cosine Rule in $ OBA_{2}$, where $ OB = R$, using value of $ BA_{2}$ found above, $ \angle OBA_{2} = 90 - A$, we have: $ OA_{2} = \sqrt {R^2 + R^2 * (\frac {sin 2B}{cos (B - C)})^2 - 2R^2 \frac {sin 2B sin A} {cos (B - C)}} = R \frac {\sqrt {(cos (B - C))^2 + (sin 2B)^2 - 2 sin 2B sin A cos (B - C) }}{ cos (B - C)}$ Note that: $ (cos(B - C))^2 + (sin 2B)^2 - 2 sin 2B sin A cos (B - C) = (cos(B - C))^2 + (sin 2B)^2 - 2 sin 2B sin (B + C) cos (B - C) = (cos(B - C))^2 + (sin 2B)^2 - sin 2B (sin 2B + sin 2C) = (cos(B - C))^2 - sin 2B sin 2C = (cos(B - C))^2 - \frac { cos (2B - 2C) - cos (2B + 2C)} {2} = (cos (B - C))^2 - \frac { (2 (cos (B - C))^2 - 1) - (2 (cos (B + C))^2 - 1) }{2} = (cos (B - C))^2 - [ (cos (B - C))^2 - (cos (B + C))^2] = (cos (B + C))^2 = (cos A)^2$, from which we have: $ OA_{2} = R \frac {cos A}{cos (B - C)}$, so $ \frac {OA_{2}}{OA_{1}} = \frac {OA_{2}}{HA_{2}} = \frac {cos A}{2 cos B cos C}$. We can prove a stronger inequality: $ \prod OA_{2} \geq \prod HA_{2}$ , from which our required inequality follows from AM-GM. This inequality reduces to the following: $ cos A cos B cos C \leq 1/8$, which is a well-known inequality that follows from Jensen applied to the concave function $ f(x) = In (cos x)$. - The fact that $ OA_{2} = \frac {R cos A}{cos (B-C)}$ is equivalant to proving that $ \angle OBA_{2} = 90 - |B-C|$, by considering Sin Rule in $ OBA_{2}$. This is equivalant to proving that $ \angle A_{2}A'' = \angle OA_{2}A''$. Perhaps there is a nice, Euclidean method of doing this, but I'm not sure how.
The QuattoMaster 6000
11.03.2009 06:34
Let $ N$ be the midpoint of $ OH$ and $ M$ be the midpoint of $ BC$. Reflect $ A$, $ B$, and $ C$ into $ N$ to form $ A'$, $ B'$, and $ C'$ respectively. Now, if $ X$ is the midpoint of $ AH$, then since $ AH = 2R\cos A = 2OM$, we have that $ HX = OM$, so $ XHMO$ is a parallelogram, and so $ X$ and $ M$ are reflections into the midpoint of $ OH$. It follows that $ X$ lies on $ B'C'$. Now, $ B'C'||BC\perp AH$, so $ HA$ is perpendicular to $ B'C'$. Furthermore, $ M$ is the midpoint of $ BC$, so $ X$ is the midpoint of $ B'C'$, so $ H$ is on the perpendicular bisector of $ B'C'$. Similarly, it is on the perpendicular bisector of $ A'B'$, so $ H$ is the circumcenter of $ \triangle A'B'C'$. Now, the reflection into $ N$ maps $ H$ to $ O$, so $ O$ is the orthocenter of $ \triangle A'B'C'$ since $ H$ was that of $ \triangle ABC$. Noitce that $ \frac {OQ_1}{OP_1} = \frac {HP_1}{OP_1}$ by the parallelogram condition. Since $ P_1$ and $ Q_1$ are reflections into $ N$, $ P_1$ lies on $ B'C'$. Thus, $ P_1$ lies on the perpendicular bisector of $ AH$, so $ AP_1 = HP_1$. Now, it suffices to show that $ \sum \frac {AP_1}{OP_1} \ge 3$. Let the altitudes of $ \triangle A'B'C'$ be $ A'J$, $ B'K$, and $ C'L$. We have that $ \frac {XH}{JO} = \frac {AX}{JO} = \frac {AP_1}{OP_1}$. It then suffices to have that $ \sum \frac {XH}{JO}\ge 3$. However, since $ A'O = AH = 2OM = 2HX$, we have that we want for\[ \sum \frac {A'O}{JO}\ge 6\iff \sum \frac {A'J}{JO}\ge 9\]Now, notice that\[ \sum \frac {JO}{A'J} = \frac {\sum [B'OC']}{[A'B'C']} = 1\]so\[ \sum \frac {JO}{AJ} \sum \frac {AJ}{JO}\ge (1 + 1 + 1)^2 = 9\]by Cauchy and we are done.
math154
23.05.2011 22:57
By simple complex numbers with $a_1,a_2,a_3$ on the unit circle, \[p_1=\frac{a_1(a_1+a_2)(a_1+a_3)}{a_1^2+a_2a_3},\; q_1=\frac{a_2a_3(a_2+a_3)}{a_1^2+a_2a_3}.\]Thus \[\frac{OQ_1^2}{OP_1^2}=\left|\frac{OQ_1^2}{OP_1^2}\right|=\left|\frac{q_1\overline{q_1}}{p_1\overline{p_1}}\right|=\left|\frac{a_1^2(a_2+a_3)^2}{(a_1+a_2)^2(a_1+a_3)^2}\right|=\frac{R^2\cdot OM_1^2}{4OM_2^2\cdot OM_3^2},\]where $M_1,M_2,M_3$ are the midpoints of $BC,CA,AB$, respectively. Since $OM_i=R\cos{A_i}$, \[\sum\frac{OQ_i}{OP_i}=\sum\frac{\cos{A_i}}{2\cos{A_{i+1}}\cos{A_{i+2}}}\ge\frac{3}{2}\sqrt[3]{\frac{1}{\cos{A_1}\cos{A_2}\cos{A_3}}}\ge3,\]where the last inequality follows from the fact that $\log\cos{x}$ is concave for $x\in(0,\pi/2)$.
brian22
17.04.2016 04:20
A few definitions:
$H_1$ is the reflection of $H$ over $A_2A_3$
$H^*_1$ is the foot of the altitude of $A_1$ onto $A_2A_3$ -- trivially, also the midpoint of $HH_1$.
$M_1$ is the midpoint of $A_2A_3$
First, we show that $H_1$, $Q_1$, and $O$ are collinear. We know that $A_1O \parallel HQ_1$, so since $H_1$ lies on $A_1H$, $\angle H_1HQ_1 = \angle H_1A_1O$. Now, it is well known that $H_1$ lies on the circumcircle of $\triangle A_1A_2A_3$, so $\angle H_1A_1O = \angle A_1H_1O$. In addition, since $H_1$ is simply the reflection of $H$ over $A_2A_3$, and $Q_1$ lies on $A_2A_3$, $\angle HH_1Q_1 = \angle H_1HQ_1 = \angle H_1A_1O$; thus, by AA similarity, $\triangle H_1HQ_1 \sim \triangle H_1A_1O$. Finally, since $A_1O \parallel HQ_1$ and trivially $H_1H \parallel H_1A_1$, we must have that $OH_1 \parallel Q_1H_1$, which implies that $O$, $Q_1$, and $H_1$ are collinear.
Now, let $R$ be the circumradius of $\triangle A_1A_2A_3$. Note that by properties of parallelograms, $OP_1 = HQ_1$, and also note that $HQ_1+OQ_1 = R$. Finally, note that $\angle HA_1O$ is the difference between $\angle OA_1A_2 = \frac{\pi}{2} - \angle A_3$ and $\angle HA_1A_2 = \frac{\pi}{2} - \angle A_2$, so it is either $A_2-A_3$ or $A_3-A_2$. Since $\cos(-x) = \cos(x)$, it won't end up mattering which one it is. Using the well-known fact that $\angle A_2OA_3 = 2 \angle A_1$, we can derive that
$$OQ_1 = \frac{OM_1}{\cos(\angle A_2 - \angle A_3)} = \frac{OA_2 \cos(\angle A_1)}{\cos(\angle A_2 - \angle A_3)} = \frac{R \cos(\angle A_1)}{\cos(\angle A_2 - \angle A_3)}$$Now, using this we can derive that
$$\frac{OQ_1}{OP_1} = \frac{OQ_1}{R-OQ_1} = \frac{R*\frac{\cos(\angle A_1)}{\cos(\angle A_2 - \angle A_3)}}{R * \frac{\cos(\angle A_2 - \angle A_3) - \cos(\angle A_1)}{\cos(\angle A_2 - \angle A_3)}} = \frac{\cos(\angle A_1)}{\cos(\angle A_2 - \angle A_3) - \cos(\angle A_1)} = \frac{\cos(\angle A_1)}{\cos(\angle A_2 - \angle A_3) + \cos(\angle A_2 + \angle A_3)} = \frac{\cos(\angle A_1)}{2\cos(\angle A_2)\cos(\angle A_3)}$$By symmetry, we have that $\frac{OQ_1}{OP_1} + \frac{OQ_2}{OP_2} + \frac{OQ_3}{OP_3} = \sum \frac{\cos(\angle A_i)}{2\cos(\angle A_{i+1})\cos(\angle A_{i+2})} \ge 3\sqrt[3]{\frac{\cos(\angle A_1)\cos(\angle A_2)\cos(\angle A_3)}{8\cos^2(\angle A_1)\cos^2(\angle A_2)\cos^2(\angle A_3)}}$
Now we prove the fact that everyone seems to be brushing aside with an unjustified "log cos is convex" -- namely, the fact that $\cos(\angle A_1)\cos(\angle A_2)\cos(\angle A_3) \le \frac{1}{8}$. To show this, first note that for $A, B \in [0, \frac{\pi}{2}]$, $\cos(2A)\cos(2B) = (\cos^2(A)-\sin^2(A))(\cos^2(B)-\sin^2(B)) = \cos^2(A)\cos^2(B)-\sin^2(A)\cos^2(B)-\cos^2(A)\sin^2(B)+\sin^2(A)\sin^2(B)$ $\le \cos^2(A)\cos^2(B)-2\sin(A)\sin(B)\cos(A)\cos(B)+\sin^2(A)\sin^2(B) = (\cos(A)\cos(B)-\sin(A)\sin(B))^2 = \cos^2(A+B)$. Using this, we have that $\cos(\angle A_1)\cos(\angle A_2)\cos(\angle A_3) \le \cos^2\left(\frac{\angle A_1+\angle A_2}{2}\right)\cos(\angle A_3) = \cos^2\left(90-\frac{\angle A_3}{2}\right)\cos(\angle A_3) = \sin^2\left(\frac{\angle A_3}{2}\right)\cos(\angle A_3)$ $ = \left(\frac{1-\cos(\angle A_3)}{2}\right)\cos(\angle A_3) = \frac{1}{8} - \frac{(\cos(\angle A_3)-\frac{1}{2})^2}{2} \le \frac{1}{8}$, as desired.
With this fact, the proof is practically done. Just note that
$$3\sqrt[3]{\frac{\cos(\angle A_1)\cos(\angle A_2)\cos(\angle A_3)}{8\cos^2(\angle A_1)\cos^2(\angle A_2)\cos^2(\angle A_3)}} = 3\sqrt[3]{\frac{1}{8\cos(\angle A_1)\cos(\angle A_2)\cos(\angle A_3)}} \ge 3\sqrt[3]{\frac{1}{8*\frac{1}{8}}} = 3$$as desired.