IMO ShortList 2001, geometry problem 2

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orl

3647 posts

orl

#1 h Sep 30, 2004, 5:39 PM • 7 Y

Y by mathmaths, r0518, Math-Ninja, Adventure10, HWenslawski, ImSh95, Mango247

Consider an acute-angled triangle $ABC$ . Let $P$ be the foot of the altitude of triangle $ABC$ issuing from the vertex $A$ , and let $O$ be the circumcenter of triangle $ABC$ . Assume that $\angle C \geq \angle B+30^{\circ}$ . Prove that $\angle A+\angle COP < 90^{\circ}$ .

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orl

3647 posts

orl

#2 h Sep 30, 2004, 5:39 PM • 6 Y

Y by r0518, Math-Ninja, alireza_32, Adventure10, ImSh95, Mango247

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

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mathmanman

1444 posts

mathmanman

#3 h Apr 15, 2006, 2:24 PM • 6 Y

Y by Math-Ninja, Not_real_name, myh2910, Adventure10, ImSh95, Mango247

Like usual, we let : $\angle A = \alpha$ , $\angle B = \beta$ and $\angle C = \gamma$ .
Then, $\angle BCO = \frac 12(\pi - \angle BOC) = \frac{\pi}2 - \frac 12(2 \alpha)$ .
Since $\triangle ABC$ is acute-angled, $O$ is inside the triangle, and $P$ is between $B$ and $C$ . So, we just have to prove that $\angle COP = \angle POC$ , or, equivalently : $PO > PC$ . Let $OC = R$ , then $b = 2R \sin{\beta}$ (thanks to the sine law).
Since $PC = b \cos{\gamma}$ , we have :
$\begin{eqnarray*}PC^2 < PO^2 &\iff& PC^2 < R^2 + PC^2 - 2Rb \cos{\gamma} \cos{(\frac{\pi}2 - \alpha)} \\ &\iff& 0 < \frac{b^2}{4 \sin^2{\beta}} - \frac{b^2 \cos{\gamma} \sin{\alpha}}{\sin{\beta}} \\ &\iff& 4 \sin{\alpha} \sin{\beta} \cos{\gamma} < 1 \\ &\iff& 2 \sin{\alpha}(\sin{(\beta + \gamma)} - \sin{(\gamma - \beta)}) <1 \\ &\Longleftarrow& 2 \sin{\alpha}(\sin{\alpha} - \sin{\frac{\pi}6}) \le 1 \\ &\Longleftarrow& 2 \sin{\alpha}\left(1 - \frac 12 \right) \le 1 \\ &\Longleftarrow& 2 - 2 \left(\frac 12 \right) \le 1 \end{eqnarray*}$
which is true.

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yetti

2643 posts

yetti

#4 h Dec 31, 2006, 2:48 AM • 1 Y

Y by Adventure10

The problem condition $C\ge B+30^\circ$ is equivalent to

$\sin(C-B) \ge \sin 30^\circ = \frac{1}{2}\ \Longleftrightarrow\ \sin C\cos B-\sin B\cos C\ge \frac{1}{2}\ \Longleftrightarrow$

$\frac{c}{2R}\cdot \frac{c^{2}+a^{2}-b^{2}}{2ca}-\frac{b}{2R}\cdot \frac{a^{2}+b^{2}-c^{2}}{2ab}\ge \frac{1}{2}\ \Longleftrightarrow\ c^{2}-b^{2}\ge aR$

The angles $A+\angle COP$ add to $90^\circ$ in an acute $\triangle ABC,$ iff the triangle $\triangle COP$ is isosceles with $\angle COP = \angle OCP,$ because it is always trus that $\angle OCP \equiv \angle OCB = 90^\circ-A$ . Thus $A+\angle COP < 90^\circ$ is equivalent to $PO > PC.$ Let M be the midpoint of BC. By Pythagorean theorem for the right triangle $\triangle OMP,$

$PO^{2}= OM^{2}+MP^{2}$

$MP = CM-CP,\ \ CM = \frac{a}{2},\ \ CP = b \cos C = \frac{a^{2}+b^{2}-c^{2}}{2a}$

$PO > PC\ \Longleftrightarrow\ OM^{2}+(CM-CP)^{2}> CP^{2}\ \Longleftrightarrow$

$OM^{2}+CM^{2}-2CM \cdot CP > 0\ \Longleftrightarrow OM^{2}+\frac{a^{2}}{4}-\frac{a^{2}+b^{2}-c^{2}}{2}> 0\ \Longleftrightarrow$

$OM^{2}+\frac{c^{2}-b^{2}}{2}> \frac{a^{2}}{4}$

The consequence $c^{2}-b^{2}\ge aR$ of the prolem condition and the trivial facts $OM > 0,$ $R > \frac{a}{2}$ for an acute triangle imply

$OM^{2}+\frac{c^{2}-b^{2}}{2}\ge OM^{2}+\frac{aR}{2}> OM^{2}+\frac{a^{2}}{4}> \frac{a^{2}}{4}$

hence $A+\angle COP < 90^\circ$ .

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BenasKikutis

3 posts

BenasKikutis

#5 h Jun 27, 2009, 7:15 PM • 3 Y

Y by Lukaluce, Adventure10, Mango247

Can anybody post solution without trigonometry, if there is one. Please

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yetti

2643 posts

yetti

#6 h Jun 27, 2009, 10:31 PM • 2 Y

Y by Adventure10, Mango247

$AP, AO$ and parallel to $BC$ through $A$ cut $(O)$ again at $X, Y, Z,$ respectively. $YZ$ cuts $BC$ at $Q.$ $XY \parallel BC,$ $YX \perp AX$ $\Longrightarrow$ $AXYZ$ is rectangle, $XY = PQ = AZ$ and $M$ is common midpoint of $BC, PQ.$ For any acute $\triangle ABC,$ $OM > 0,$ $R > \frac {BC}{2}$ and

$\angle COP < \angle OCP = 90^\circ - \angle A$ $\Longleftrightarrow$ $CP^2 < OP^2$ $\Longleftrightarrow$ $\frac {(BC - PQ)^2}{4} < OM^2 + \frac {PQ^2}{4}$ $\Longleftrightarrow$ $OM^2 + \frac {BC \cdot PQ}{2} > \frac {BC^2}{4}$

From $\angle XAY = \angle C - \angle B \ge 30^\circ$ $\Longrightarrow$ $PQ = XY \ge \frac {AY}{2} = R.$ Therefore,

$OM^2 + \frac {BC \cdot PQ}{2} \ge OM^2 + \frac {BC \cdot R}{2} > \frac {BC^2}{4}$ $\Longrightarrow$ $CP < OP$ $\Longrightarrow$ $\angle COP + \angle A < 90^\circ.$

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math154

4302 posts

math154

#7 h Jul 18, 2010, 7:09 PM • 2 Y

Y by Adventure10, Mango247

Blah here's my silly way to show $OP>CP$ .

From the power of point $P$ we see that $R^2-OP^2=BP\cdot CP$ , so it is equivalent to show that
$OP>\frac{R^2-OP^2}{BP}=CP,$ or by the Law of Cosines on $\angle{BPO}$ , that
$\begin{align*} OP\cdot BP+OP^2 &> R^2 = BO^2 = BP^2+OP^2-2BP\cdot OP\cos\angle{BPO}\\ \Longleftrightarrow OP+OP(2\cos\angle{BPO}) &> BP. \end{align*}$ But by the Law of Sines on $\triangle{APO}$ , we have
$2\cos\angle{BPO} = 2\sin\angle{APO} = 2\frac{AO\sin(C-B)}{OP} \ge \frac{AO}{OP}.$ Hence
$OP+OP(2\cos\angle{BPO}) \ge OP+AO = OP+BO > BP$ by the Triangle Inequality, so we're done.

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golem24

92 posts

golem24

#8 h Jun 5, 2012, 7:04 PM • 3 Y

Y by samrocksnature, Nafis_Noor, Adventure10

Draw $OK\perp AP$ and $OD\perp BC$ .Obviously $KOPD$ is a rectangle.
$AO=CO$ means $\triangle ABC$ is isosceles. So $\angle CAO=\frac{1}{2}(\pi-\angle AOC)=\frac{\pi}{2}-\angle B$

$[asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11.74cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 7.44, ymin = -2.64, ymax = 6.3; /* image dimensions */ /* draw figures */ draw(circle((1.82,1.42), 3.76)); draw((0.29,4.86)--(-1.59,-0.16)); draw((-1.59,-0.16)--(5.21,-0.21)); draw((5.21,-0.21)--(0.29,4.86)); draw((0.29,4.86)--(0.27,-0.17)); draw((1.82,1.42)--(0.28,1.44)); draw((1.82,1.42)--(-1.59,-0.16)); draw((1.82,1.42)--(0.27,-0.17)); draw((1.82,1.42)--(1.8,-0.19)); /* dots and labels */ dot((1.82,1.42),dotstyle); label("$O$", (2,1.46), NE * labelscalefactor); dot((0.29,4.86),dotstyle); label("$A$", (0.38,4.98), NE * labelscalefactor); dot((-1.59,-0.16),dotstyle); label("$C$", (-1.42,-0.38), NE * labelscalefactor); dot((5.21,-0.21),dotstyle); label("$B$", (4.78,-0.52), NE * labelscalefactor); dot((0.27,-0.17),dotstyle); label("$P$", (0.24,-0.52), NE * labelscalefactor); dot((0.28,1.44),dotstyle); label("$H$", (0.36,1.56), NE * labelscalefactor); dot((1.8,-0.19),dotstyle); label("$D$", (1.74,-0.48), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

$\angle CAP=\frac{\pi}{2}-\angle C$ , so angle $PAO=\angle CAO-\angle CAP=\angle C-\angle B\ge 30^\circ$ .

So we have $KO\ge \frac{1}{2}AO$ (You may verify this using sine rule as well).As $KOPD$ is a rectangle $KO=PD>\frac{1}{2}AO=\frac{1}{2}CO>\frac{1}{2}CD$ . So $PC<PD<OP\implies \angle COP<\angle PCO=\angle OCD \dots (*)$ .

We now have $\angle COP+ \angle A=\angle COP+\angle COD<\angle OCD+\angle COD=90^\circ$ .

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hyperspace.rulz

287 posts

hyperspace.rulz

#9 h Apr 9, 2013, 1:43 PM • 2 Y

Y by Adventure10, Mango247

Another approach that utilises power of a point:

As in all the other solutions here, we establish that we need to prove $OP>CP$ . This in turn is equivalent, obviously, to:

$OP^2>CP^2$
$\Leftrightarrow R^2-PC \cdot PB>CP^2$

This is because of Power of a Point. We have that this is in turn equivalent to:

$R^2>CP^2+PC \cdot PB=CP(CP+PB)=CP \cdot CB$

Now note that $PB=AB\cos B=2R\sin C\cos B$ , and similarly $PC=2R\sin B\cos C$ . It follows that $PB-PC=2R\sin(C-B) \geq R$ , where the last step follows from the problem's condition. Since $PB+PC=BC$ , we deduce that $PB \geq \frac{R+BC}{2}$ , and hence $PC \leq \frac{BC-R}{2}$ . So we have:

$CP \cdot CB \leq \frac{BC(BC-R)}{2}$ , so it suffices to show that:

$\frac{BC(BC-R)}{2}<R^2$
$\Leftrightarrow \frac{a(a-R)}{2}<R^2$
where $BC=a$ . Bearing in mind that $a<2R$ , we have:

$a-R<R$
$\Leftrightarrow (a-R)(a+R)<R(a+R)$
$\Leftrightarrow a^2-R^2<aR+R^2$
$\Leftrightarrow a^2-aR<2R^2$
$\Leftrightarrow a(a-R)<2R^2$
$\Leftrightarrow \frac{a(a-R)}{2}<R^2$

as desired.

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JSGandora

4216 posts

JSGandora

#10 h May 11, 2013, 4:57 PM • 2 Y

Y by Taco12, Adventure10

Extend $CO$ to meet the circumcircle of $ABC$ at $D$ . Then $\angle DAB+\angle A=90^\circ$ and we want to prove that
$\angle A+\angle COP<90^\circ=\angle DAB+\angle A\Longleftrightarrow \angle COP<\angle DAB.$
We also have $\angle DAB=\angle BCD$ so what we want to prove is equivalent to proving
$\angle COP<\angle BCD=\angle PCO$
which is also equivalent to proving $OP>CP$ .

We have
$\begin{align*} OP&>CP\\ \Longleftrightarrow OP^2&>CP^2 \\ \Longleftrightarrow CP^2+OC^2-2(CO)(CP)\cos\angle PCO&>CP^2\\ \Longleftrightarrow OC^2&>2(OC)(CP)\cos\angle PCO \\ \Longleftrightarrow OC&>2(CP)\cos \angle PCO \end{align*}$
Now let $\angle B=\beta$ and $\angle C=\gamma$ , then $\angle PCO=\beta+\gamma-90$ . Also, $PC=b\cos \gamma$ and $OC=\frac{b}{2\sin \beta}$ . Now what we want to prove is
$\frac {b}{2\sin \beta}>2b\cos \gamma\cos(\beta+\gamma-90)=2b\cos\gamma\sin(\beta+\gamma)$
or
$1>4\sin\beta\cos\gamma\sin(\beta+\gamma).$
By the condition, $\gamma\geq \beta+30$ and the triangle is acute so $\cos\gamma<\cos(\beta+30)$ . Then
$4\sin\beta\cos\gamma\sin(\beta+\gamma)<4\sin\beta\cos(\beta+30)=2\sin\beta(\cos\beta\sqrt3-\sin\beta).$
Taking the derivative of the RHS we get
$2(\cos \beta(\cos\beta\sqrt3-\sin\beta)+\sin\beta(-\sin\beta\sqrt3-\cos\beta)) \\ =2(\sqrt{3}(\cos^2\beta-\sin^2\beta-2\sin\beta\cos\beta))=2(\sqrt3\cos2\beta-\sin2\beta)$
We find the critical points:
$\begin{align*} 2(\sqrt3\cos2\beta-\sin2\beta)&=0 \\ \sqrt{3}\cos 2\beta=\sin2\beta \end{align*}$
Since $\beta\neq 45^\circ$ , the LHS is not equal to 0 so we divide
$\tan 2\beta=\sqrt{3}$
and the only solution to this is such that $0^\circ<\beta<60^\circ$ $2\beta=60\circ\implies \beta=30^\circ$ . Thus the critical points are $\beta=\{0^\circ,30^\circ,60^circ\}$ . Plugging the two critical points in, we have
$2\sin\beta(\cos\beta\sqrt3-\sin\beta)=0$
for $\beta=0^\circ$ and $\beta=60^\circ$ and
$2\sin\beta(\cos\beta\sqrt3-\sin\beta)=1$
when $\beta=30^\circ$ therefore the maximum is achieved when $\beta=30^\circ$ however that means $\gamma =90^\circ$ which is impossible since the triangle is acute so therefore the maximum is not achievable and thus
$4\sin\beta\cos\gamma\sin(\beta+\gamma)<1$
as desired. $\blacksquare$

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sayantanchakraborty

505 posts

sayantanchakraborty

#11 h Apr 22, 2014, 5:52 PM • 1 Y

Y by Adventure10

Outlines of my proof

Let $\angle{COP}=\theta$ .Easy to show that its acute.
Sine rule in $\triangle{COP}$ gives
$\frac{R}{CP}=\frac{cos(A-\theta)}{sin\theta} \Rightarrow cosA cot\theta+sinA=\frac{1}{2sinBcosC}=\frac{1}{sinA-sin(C-B)} \ge \frac{1}{sinA-\frac{1}{2}}=\frac{2}{2sinA-1}$ .

From here we get the upper bound of $\tan\theta$ .Now it is easy(using some elementary calcuations) to show that $tanA tan\theta <1$ so that $tan{A+\theta}>0$ and the result follows...

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CanVQ

83 posts

CanVQ

#12 h Apr 23, 2014, 3:12 PM • 2 Y

Y by Adventure10, Mango247

orl wrote:

Consider an acute-angled triangle $ABC$ . Let $P$ be the foot of the altitude of triangle $ABC$ issuing from the vertex $A$ , and let $O$ be the circumcenter of triangle $ABC$ . Assume that $\angle C \geq \angle B+30^{\circ}$ . Prove that $\angle A+\angle COP < 90^{\circ}$ .

My solution:
Click to reveal hidden text
Let $D$ be the foot of the altitude of $\triangle ABC$ issuing from $B.$ It is known that $DP \perp OC$ and $\triangle CPD \sim \triangle CAP.$

Let $E=OC \cap DP,$ we have $\cos A=\cos \angle CPE=\frac{PE}{PC}$ and $\sin \angle COP=\frac{PE}{OP}.$ Therefore, $\angle A+\angle COP<90^{\circ} \Leftrightarrow \sin \angle COP<\sin (90^{\circ}-\angle A)=\cos A\Leftrightarrow OP>CP.$ Now, let us compute the lengths of $OP$ and $CP.$ Let $M$ be the midpoint of $BD$ and $R$ be the radius of $(O),$ we have $CP=AC\cdot \cos \angle ACP=2R\cdot \sin B\cos C$ and $CM=OC\cdot \sin COM=R\cdot \sin A,\quad OM=OC\cdot \cos \angle COM=R\cdot \cos A.$ Therefore, $\begin{aligned} OP^2&=OM^2+PM^2=OM^2+(CM-CP)^2\\ &=R^2\big[ \cos^2A+(\sin A- 2\sin B\cos C)^2\big].\end{aligned}$ It suffices to prove that $4\sin^2B\cos^2C<\cos^2A+(\sin A-2\sin B\cos C)^2,$ or $1>4\sin A\sin B\cos C.$ We have $\begin{aligned} 4\sin A\sin B\cos C&=2\sin A \big[ \sin A-\sin (C-B)\big] \le 2\sin A(\sin A-\sin 30^{\circ})\\ &=2\sin^2A-\sin A =1+(\sin A-1)(2\sin A+1)<1.\end{aligned}$ The proof is done.

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guldam

31 posts

guldam

#13 h Jan 13, 2015, 3:07 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

I hope this solution would not be duplicate with any other presented above.

Since $\angle PCO = 90 - \angle A$ it is well noted that to prove $PO > PC$ is enough.

It is also well-known that $\angle OAP = \angle C - \angle B \ge 30$

Let point $M$ to be the midpoint of $BC$ then

$MP = R\sin{\angle OAP} \>= R \sin{30} = \frac{R}{2}$

since $30\le \angle OAP < \angle A \le 90$

$R < a = 2R \sin{A} < 2R$

therefore,

$PC = MC - MP \le \frac{a-R}{2} < \frac{R}{2} \le MP < PO$ $Q.E.D$

the last inequality hold since $PO$ is the hypotenuse of the right triangle $\triangle PMO$

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TelvCohl

2312 posts

TelvCohl

#14 h Jan 13, 2015, 5:31 PM • 3 Y

Y by jonathan3951, guptaamitu1, Adventure10

My solution:

Let $A'$ be the reflection of $A$ in the perpendicular bisector of $BC$ .
Let $M, N$ be the midpoint of $BC, AA'$ , respectively .

Since $\angle AOA'=2(\angle ACB-\angle CBA) \geq 60^{\circ}$ ,
so we get $AA' \geq OA \Longrightarrow MP=AN \geq \tfrac{1}{2} OA$ ,
hence $CP=MC-MP<OC-MP \leq \tfrac{1}{2} OA \leq MP \Longrightarrow \angle COP<\angle OCB$
$\Longrightarrow \angle BAC+\angle COP=\angle MOC+\angle COP<\angle MOC+\angle OCB=90^{\circ}$

Q.E.D

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va2010

1276 posts

va2010

#15 h Mar 30, 2016, 1:21 AM • 3 Y

Y by Adventure10, Mango247, shafikbara48593762

Here's a solution, from working with swimmer:

We just have to show $PO > PC$ . But, using the following diagram:

https://hostr.co/file/J105WwHiuEme/IMOGEO.png

we can see that $PO > FO > EF > CP$ , so we are done.

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YaoAOPS

1541 posts

YaoAOPS

#38 h Aug 16, 2022, 7:50 PM

Y by

Storage

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bobthegod78

2982 posts

bobthegod78

#39 h Aug 17, 2022, 12:14 AM

Y by

Claim: $R^2 > ab \cos C$ .
Proof: By the extended law of sines, $R^2 = \frac{a}{2 \sin A} \cdot \frac{b}{2\sin B}$ . We can divide out $ab \cos C$ to get $\frac{R^2}{ab \cos C} = \frac{1}{4\sin A \sin B \cos C}$ . But $\sin B \cos C = \frac 12 ( \sin(B+C) + \sin(B-C)) = \frac 12(\sin A - \sin (C-B))$ , so $\frac{R^2}{ab \cos C} = \frac{1}{2 \sin A (\sin A - \sin(C-B))}.$ But notice that $2\sin A (\sin A - \sin (C-B))$ is maximized when $\sin A$ is maximized and $\sin(C-B)$ is minimized, and since $A<90$ , $\sin A < 1$ and since $C-B \geq 30,$ $\sin(C-B) \geq 1/2$ , so $2\sin A ( \sin A - \sin (C-B)) < 1$ , or $\frac{R^2}{ab \cos C} > 1$ , implying $R^2 > ab\cos C$ , as desired.

Claim: $PO>CP$ .
Proof: We find that $CP = b \cos C$ and $PO = \sqrt{(\frac a2 - b \cos C)^2 + R^2 - (\frac a2)^2} = \sqrt{b^2 \cos^2 C + R^2 - ab \cos C}.$ Since $R^2 - ab\cos C$ , $\sqrt{b^2 \cos^2 C + R^2 - ab \cos C} > \sqrt{b^2 \cos^2 C} = b \cos C$ , so $PO>CP$ .

Since $PO>CP$ , then $90 - \angle A =\angle PCO > \angle COP$ , but then $\angle A + \angle COP < \angle A + \angle PCO = 90$ , so we are done.

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math_comb01

662 posts

math_comb01

#40 h Jun 4, 2023, 4:45 PM

Y by

Though similar to others, I find this trig bash very less bashy
The key claim is as follows,
Claim : $R^2 \geq ab cos{c}$
Proof : We first prove a lemma
Lemma: $sin(a)sin(b)cos(c) \leq \frac{1}{4}$ equivalently $sin(a)(sin(b+c)-sin(c-b)) \leq \frac{1}{2}$ $sin(a)(2sin(a)+1) \leq 1$ $(sin{a}-1)(2sin{a}+1) \leq 0$ which is true
By sine rule the claim immediately follows, and by cosine rule in $\triangle OCP$ the problem gets finished.
EDIT: just noticed this is identical to mathmanman's soln lol

This post has been edited 4 times. Last edited by math_comb01, Jun 4, 2023, 4:51 PM

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lrjr24

967 posts

lrjr24

#41 h Jun 16, 2023, 5:29 PM

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Note that $A + \angle COP < 90 \equiv \angle COP < 90 - A = \angle OCP \equiv CP<OP$ . We now bash. Note that $OP>CP$ is equivalent to $R^2\cos^2a + \left(\frac{a}{2}-b \cos C\right)^2 > b^2\cos^2C \equiv ab \cos C < R^2\cos^2A+\frac{a^2}{4} = R^2 = \frac{(abc)^2}{16 \cdot [ABC]^2} = \frac{a^2b^2c^2}{4 \cdot bc \sin A \cdot ac \sin B} \equiv \sin A \sin B \cos C < \frac{1}{4}.$ Switch to radians. If $B \le \frac{\pi}{6}$ , we need to prove $\sin B \sin \left(B+\frac{\pi}{6}\right) = f(B) \le 1/4$ . We use calculus to see $f'(B) = \sin\left( 2B+\frac{\pi}{6} \right) > 0$ . Thus we only need to check $B=\frac{\pi}{6}$ , which is obvious. We now note that if we fix $B$ and vary $C$ from $B+\frac{\pi}{6}$ , $\cos C$ will decrease and $\sin A$ will decrease. Thus we want $C=B+\frac{\pi}{6}$ . Thus we need to prove $g(x) = \sin\left(2x+\frac{\pi}{6}\right) \sin x \cos \left(x+\frac{\pi}{6}\right) < \frac{1}{4}$ for $\frac{\pi}{6} < x < \frac{\pi}{3}$ . We use calculus again. Note that $g'(x) = \cos\left(2x+\frac{\pi}{6}\right) \sin x \cos \left(x+\frac{\pi}{6}\right) + \sin\left(2x+\frac{\pi}{6}\right) \cos \left(2x+\frac{\pi}{6}\right) = \cos\left(2x+\frac{\pi}{6}\right) \left( \sin x \cos \left( x + \frac{\pi}{6} \right) + \sin \left( 2x +\frac{\pi}{6} \right) \right) < 0$ , so we just need to check $x = \frac{\pi}{6}$ which is obvious and we're done.

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smileapple

1010 posts

smileapple

#42 h Jul 27, 2023, 6:54 AM

Y by

$[asy] unitsize(2.8cm); pair A,B,C,O,P; A=(0.4,0.9165151389912); B=(-0.8,-0.6); C=(0.8,-0.6); O=(0,0); P=(0.4,-0.6); draw(circle((0,0),1)); draw(A--B--C--cycle); draw(O--C); draw(B--O--P--A); label("$A$",A,NE); label("$B$",B,SW); label("$C$",C,SE); label("$O$",O,N); label("$P$",P,S); [/asy]$
Note that $\angle BOC=2\angle A$ due to the Inscribed Angle Theorem. Since $\triangle BOC$ is isosceles with $BO=OC$ , it follows that
$\begin{align*} \angle OBC&=\angle OCB\\ &=\frac{180^{\circ}-\angle BOC}2\\ &=\frac{180^{\circ}-2\angle A}2\\ &=90^{\circ}-\angle A. \end{align*}$ Thus, proving that $\angle A+\angle COP<90^{\circ}$ is equivalent to proving that $\angle COP<90^{\circ}-\angle A=\angle OCP.$ Applying the Law of Sines, we find that $\frac{OP}{\angle OCP}=\frac{PC}{\angle COP}$ , so that $\frac{OP}{PC}=\frac{\sin\angle OCP}{\sin\angle COP}.$ If $OP<PC$ , then $\sin\angle OCP<\sin \angle COP$ and $\angle COP<\angle OCP$ , implying that $\angle OCP+\angle COP>180^{\circ}$ , a contradiction. It therefore follows that having $\angle A+\angle COP<90^{\circ}$ is equivalent to $\angle COP<\angle OCP$ , which in turn is the same as having $PC<OP$ , as proven above. Thus, it suffices to show that $PC<OP.$

Defining the side lengths $(BC,CA,AB)=(a,b,c)$ , as usual, we note that $PC=b\cos\angle C$ by analysis on $\triangle ACP.$ Moreover, applying the Law of Cosines on $\triangle OCP$ gives
$\begin{align*} OP^2&=OC^2+CP^2\\ &=OC^2+CP^2-2OC\cdot CP\cos\angle OCP\\ &=PC^2+R^2-2R(b\cos\angle C)\cos(90^{\circ}-\angle A)\\ &=PC^2+R^2-2Rb\cos\angle C\sin\angle A, \end{align*}$ where $R=OA=OB=OC$ is the circumradius of $\angle ABC.$

Now, note that $R=\frac{b}{2\sin\angle B}$ by the Law of Sines, it follows that
$\begin{align*} \frac{2(OP^2-PC^2)\sin\angle B}{bR}&=\frac{2R\sin\angle B-4b\cos\angle C\sin\angle B\sin\angle A}b\\ &=\frac{b-4b\cos\angle C\sin\angle B\sin\angle A}b\\ &=1-4b\cos\angle C\sin\angle B\sin\angle A. \end{align*}$ Since $\frac{2\sin\angle B}{bR}$ is clearly positive, we see that $\frac{2(OP^2-PC^2)\sin\angle B}{bR}=1-4b\cos\angle C\sin\angle B\sin\angle A$ and $OP^2-PC^2$ share the same sign. Thus, to prove that $PC<OP$ , it suffices to show that $\cos\angle C\sin\angle B\sin\angle A<\frac14.$
\paragraph{}Note that $\sin\angle A<1$ due to the acuity of $\angle ABC$ ; in addition, the fact that $\angle C\ge\angle B+30^{\circ}$ implies that $\cos\angle C\le\cos(\angle B+30^{\circ})$ as the cosine function is decreasing on the range $[0^{\circ},180^{\circ}].$ Therefore
$\begin{align*} \cos\angle C\sin\angle B\sin\angle A&<\cos\angle C\sin\angle B\\ &\le\cos(\angle B+30^{\circ})\sin\angle B\\ &=\sin(90^{\circ}-(\angle B+30^{\circ})\sin\angle B\\ &=\sin(60^{\circ}-\angle B)\sin\angle B. \end{align*}$ Applying the Sum-to-Product Identity for products of sines, we find that
$\begin{align*} \cos\angle C\sin\angle B\sin\angle A&<\sin(60^{\circ}-\angle B)\sin\angle B\\ &=\frac{\cos((60^{\circ}-\angle B)-\angle B)-\cos((60^{\circ}-\angle B)+\angle B)}2\\ &=\frac{\cos(60^{\circ}-2\angle B)-\cos60^\circ}2\\ &\le\frac{1-\frac12}2\\ &=\frac14, \end{align*}$ as desired. $\blacksquare$

This post has been edited 1 time. Last edited by smileapple, Jul 27, 2023, 6:54 AM

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huashiliao2020

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huashiliao2020

#43 h Aug 30, 2023, 10:30 PM

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Remark: Reasonably easy problem, if you ask me. In any diagram, it looks to be the case that COP+BAC<OCP+BAC=90, finishing, so it suffices to prove CP<OP. There's also a sort of degree of "equality" even though the thing we want is strict, and even though this is geometry: When I saw this, I thought of "What if BAC approaches 90 degrees? What becomes degenerate?" Well, O lies on CP, and any right angles dropped from O would lie on BCP, so weaker inequalities like leg<hypotenuse are viable.

Indeed, construct Q=foot from O on BC, R=foot from O on AP; $ARO=90,RAO=BAP-BAO=90-B-(90-C)=C-B\ge 30\implies PQ=OR\ge\frac12AO=\frac12CO>\frac12CQ\implies OP>PQ>CP,$ as desired. $\blacksquare$

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Duck123

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Duck123

#44 h Dec 21, 2023, 11:22 AM • 1 Y

Y by CT17

First post:

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shendrew7

799 posts

shendrew7

#45 h Dec 27, 2023, 12:32 AM

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Denote the projections from $O$ onto $BC$ and $AP$ as $M$ and $N$ , respectively. Then
$\angle OAN = C-B \ge 30 \implies \sin OAN \ge \frac 12.$
Thus we get that
$\frac 12 \leq \frac{NO}{AO} = \frac{PM}{CO} < \frac{PM}{CM} \implies PC < PM < PO.$
As a result, we have $\angle COP < \angle PCO = 90-A$ , giving the desired. $\blacksquare$

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cursed_tangent1434

655 posts

cursed_tangent1434

#46 h Dec 27, 2023, 12:46 AM • 1 Y

Y by GeoKing

Let $D$ be the second intersection of $CO$ and $(ABC)$ . Now, notice that
$\begin{align*} \angle CAB + \angle COP &< 90^\circ\\ \angle COP &< 90-\angle CAB\\ \angle COP &< \angle DAB \\ \angle COP &< \angle DCB = \angle OCP \end{align*}$ Thus, the required conclusion rewrites to showing $\angle COP < \angle OCP$ or in turn $PC < OP$ .
Now, let $X = \overline{OC} \cap \overline{AP}$ . First, notice that
$PC < XC \text{ ($\angle XPC= 90^\circ$ and thus $\angle PXC < 90$)}$ Next by Ratio Lemma,
$\begin{align*} \frac{OX}{XC} &= \frac{AO}{AC} \cdot \frac{\sin{\angle OAP}}{\sin{\angle PAC}}\\ OX &= \left( \frac{AO}{AC}\cdot \frac{\sin{\angle OAP}}{\sin{\angle PAC}} \right) XC. \end{align*}$ Now,
$\begin{align*} \frac{AO}{AC}\cdot \frac{\sin{\angle OAP}}{\sin{\angle PAC}} &= \frac{\sin{\angle OCA}}{\sin{\angle AOC}}\cdot \frac{\sin{\angle OAP}}{\sin{\angle PAC}} \\ &= \frac{\sin{90-\angle B}}{\sin{\angle B}} \cdot \frac{\sin{\angle B + \angle C}}{\sin{90-\angle C}}\\ &= \left(\frac{BP}{AB}\cdot\frac{AC}{PC}\right) \left(\frac{\frac{AP}{AB}\cdot\frac{PC}{AC}+\frac{AP}{AC}\cdot\frac{BP}{AB}}{2\cdot \frac{AP}{AB}\cdot\frac{BP}{AB}}\right)\\ &= \frac{PC+BP}{2PC} > 1 \end{align*}$ since $A$ is closer to $C$ than $B$ ( $\angle BCA > \angle ABC$ ) and $P$ is the foot of the altitude. Further, $\angle OXP = 90 + \angle OCP > \angle OPA \text{ (this angle is strictly less than $90^\circ$)}$ So, $OX< OP$ as well. Thus, we have that
$PC < XC < OX < OP$ and thus, $PC < OP$ as needed.

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HamstPan38825

8872 posts

HamstPan38825

#47 h Feb 27, 2024, 2:20 AM

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Notice that because $\angle OCP = 90^\circ - \angle A$ , it suffices to show that $\angle COP < \angle OCP$ , or $CP < OP$ . Using the Law of Cosines, $CP^2 < OP^2 = CP^2 + OC^2 - 2 (OC) (CP)( \sin A) \iff 1 < 4 \sin B \cos C \sin A$ as $CP = 2R \sin B \cos C$ . On the other hand, note that $2\sin B \cos C \sin A < 2\sin B \cos C = \sin(B+C) - \sin(C-B) \leq \sin(B+C) -\frac 12 \leq \frac 12$ because $C-B\geq 30^\circ$ , as needed.

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Danielzh

492 posts

Danielzh

#48 h Jul 27, 2024, 3:45 PM

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Motivation
Let's take a look at the equality case of $\angle{A}+\angle{COP}=90^{\circ}$ . We construct the antipode of $B$ as $D$ . Notice that $\angle{A}+\angle{DAB}=90^{\circ}$ and $\angle{DAB}=\angle{DBC}$ , so the equality case holds when $CP=PO$ . We want to show that $CP<PO$ for all such $\triangle{ABC}$ . Additionally, how can be use the fact that $\angle{C} \ge \angle{B} + 30^{\circ}$ ? How can we extract the difference between the $\angle{B}$ and $\angle{C}$ ?

Solution
Let $H$ denote the orthocenter of $\triangle{ABC}$ . By properties of an orthocenter, denote the reflection of $H$ over $\overline{BC}$ , which lies on the circumcircle of $\triangle{ABC}$ , as $E$ . Hence, $90^{\circ}-\angle{C}=\angle{HBP}=\angle{HBE}$ and $\angle{ABE}=90^{\circ}+\angle{B}-\angle{C} \le 60^{\circ}$ Denote the projection from $O$ onto $\overline{BC}$ has $M$ . Examining $\triangle{AOE}$ (note that $\angle{AOE} \le 120^{\circ}$ ), we get that $MP \ge \frac{R}{2}$ , so $CP \le \frac{R}{2}$ .
By the inequalities
$\begin{align*} CP+PO > R \\ \frac{R}{2} \ge CP \\ \end{align*}$ we get that $PO > \frac{R}{2} \ge {PC}$ , which completes the proof.

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EpicBird08

1756 posts

EpicBird08

#49 h Aug 14, 2024, 12:43 AM

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What an instructive trigonometry problem.

The problem is equivalent to showing that $\angle COP < 90^\circ - \angle A.$ Notice that $\angle BCO = 90^\circ - A,$ so by considering $\triangle COP,$ it suffices to show that $PC < OP.$ Squaring both sides, this becomes $PC^2 < OP^2.$ However, if we let $R$ be the circumradius of $\triangle ABC,$ by Power of a Point we get $OP^2 = R^2 - BP \cdot PC.$ Thus it amounts to show that $PC^2 < R^2 - BP \cdot PC \leftrightarrow PC(BP + PC) = CP \cdot CB < R^2.$ By the Law of Sines, we have that $BC = 2R \sin \angle A$ and $AC = 2R \sin \angle B,$ so $PC = AC \cos \angle C = 2R \sin \angle B \cos \angle C.$ Plugging in, this becomes $4R^2 \sin \angle A \sin \angle B \cos \angle C < R^2,$ or after dividing both sides by $R^2,$ $4 \sin \angle A \sin \angle B \cos \angle C < 1.$ By the product-to-sum identities, $\sin \angle B \cos \angle C = \frac{\sin (\angle C + \angle B) - \sin (\angle C - \angle B)}{2} = \frac{\sin \angle A - \sin (\angle C - \angle B)}{2},$ and plugging this in, we get $2 \sin \angle A (\sin \angle A - \sin (\angle C - \angle B)) < 1.$ We are given that $\angle C - \angle B \ge 30^\circ,$ so $\sin (\angle C - \angle B) \ge \sin 30^\circ = \frac{1}{2}.$ Therefore, it suffices to show that $2 \sin \angle A (\sin \angle A - 0.5) = 2 \sin^2 \angle A - \sin \angle A < 1,$ or upon rearrangement, $2 \sin^2 \angle A - \sin \angle A - 1 = (2 \sin \angle A + 1)(\sin \angle A - 1) < 0.$ The factor on the left is positive while the factor on the right is negative, so the entire expression is negative. Since all our steps are reversible, we are done.

This post has been edited 1 time. Last edited by EpicBird08, Aug 14, 2024, 3:28 AM

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ezpotd

1310 posts

ezpotd

#50 h Oct 16, 2024, 3:00 AM

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This is equivalent to proving $OP > PC$ , which is equivalent to $PM \ge PC$ or $\frac{PB}{CP} \ge 3$ . Observe this is equivalent to proving $\frac{\tan B + 30}{\tan B} \ge 3$ , which is equivalent to $\frac{1 + \frac{1}{\sqrt{3} \tan B}}{1 - \frac{\tan B}{\sqrt{3}}} \ge 3$ which is equivalent to $3 \tan B + \frac{1}{\tan B} \ge 2\sqrt{3}$ , which is true by AM-GM.

This post has been edited 1 time. Last edited by ezpotd, Oct 16, 2024, 3:03 AM

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legogubbe

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legogubbe

#51 h Apr 6, 2025, 2:57 PM

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Let $Q$ be the intersection point of $OC$ and $AP$ . We find $\angle OAP = C-B \geq 30^{\circ}$ . Moreover $\angle PAC=90^{\circ}-C$ . The Sine rule applied to triangles $AOQ$ and $CAQ$ imply $\frac{OQ}{\sin(C-B)} = \frac{R}{\sin \angle OQA}$ and $\frac{QC}{\sin (90^{\circ}-C)} = \frac{2R \sin B}{\sin \angle CQA}$ , respectively, where $R$ is the circumradius of triangle $ABC$ . Hence
$\frac{OQ}{QC} = \frac{\sin(C-B)}{2\sin B \cos C}>1,$ because $\sin (C-B)>2\sin B \cos C \Leftrightarrow \sin(C-B)> \sin(B-C) + \sin(B+C) \Leftrightarrow 2 \sin (C-B)>\sin(B+C)$ which is true since $2 \sin (C-B) \geq 2 \sin 30^{\circ} = 1 > \sin (B+C)$ .

Lastly, from the Sine rule applied to triangles $OPQ$ and $PQC$ we get $\frac{OP}{\sin \angle OQP} = \frac{OQ}{\sin \angle OPQ}$ and $\frac{PC}{\sin \angle PQC} = \frac{QC}{\sin \angle CPQ}$ where
$\frac{OP}{PC} = \frac{\sin \angle CPQ}{\sin \angle OPQ} \cdot \frac{OQ}{QC}.$ As $\triangle ABC$ is acute-angled, $O$ is an interior point of $\triangle ABC$ . In fact, $\angle OPC < 180^{\circ}$ so $\angle OPQ < \angle CPQ$ . In other words, the two ratios in the equation above must be $>1$ and consequently, $OP>PC$ . We are done.
$[asy] import geometry; size(8cm); point A=(6,14), B=(-8,0), C=(9,0); triangle t=triangle(A,B,C); point O=circumcenter(t), P=projection(line(B,C))*A; point Q=intersectionpoint(line(O,C),line(A,P)); fill(A--B--C--cycle,cyan+white+white+white); draw(t,blue); draw(A--P^^O--A^^O--B^^O--C^^O--P,blue); markrightangle(A,P,C); dot("$A$",A,N); dot("$B$",B,S); dot("$C$",C,S); dot("$O$",O,N+W); dot("$P$",P,S); dot("$Q$",Q,N+E); [/asy]$

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Nari_Tom

117 posts

Nari_Tom

#52 h Apr 11, 2025, 6:54 AM

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This problem was depressing one.

Since problem is equivalent to showing: $PO^2>PC^2$ $\iff$ $R^2-BP*PC>PC^2$ $\iff$ $R^2>CP*CB$ . From here i could not found any nice ideas, so i decided to bash out like i would do on the actual contest.

So trigonometric form of this inequality is: $\frac{1}{4}> sin(\alpha)sin(\beta)cos(\gamma)$ , where $\gamma \ge \beta+30^{\circ}$ . Let's fix $\alpha$ . Then maximum value occurs when $\beta$ and $\gamma$ are close as possible. So let's assume: $\gamma=\beta+30^{\circ}$ and $\alpha=150^{\circ}-2 \beta$ . Since $\triangle ABC$ is acute we have: $30^{\circ} < \beta < 60^{\circ}$ .

Notice that $sin(150)-2 \beta$ is decreasing. Also note that $f(\beta)=sin(\beta)cos(30^{\circ}+\beta)$ is also a decreasing since: $f'(\beta)=cos(\beta)cos(30+\beta)-sin(\beta)sin(30+\beta)=cos(2 \beta+30)<0$ in the interval.

So $\frac{1}{4}=sin(30)sin(90)cos(60)>sin(b)cos(\beta+30)sin(150-2 \beta) \ge sin(\beta)sin(\alpha)cos(\gamma)$ .

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