Problem 1: Triangle triviality

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

7579 posts

#1 h Jul 12, 2006, 12:04 PM • 16 Y

Y by Davi-8191, samrocksnature, mathematicsy, Adventure10, jhu08, mathlearner2357, megarnie, son7, HWenslawski, Derpy_Creeper, michaelwenquan, jmiao, Mango247, Rounak_iitr, and 2 other users

Let $ABC$ be triangle with incenter $I$ . A point $P$ in the interior of the triangle satisfies $\angle PBA+\angle PCA = \angle PBC+\angle PCB.$ Show that $AP \geq AI$ , and that equality holds if and only if $P=I$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

shobber

3498 posts

shobber

#2 h Jul 12, 2006, 12:27 PM • 18 Y

Y by NewAlbionAcademy, JasperL, anser, ks_789, The_Outsider, cadaeibf, Adventure10, jhu08, son7, HWenslawski, michaelwenquan, jmiao, ehuseyinyigit, miloss.delfosontop, and 4 other users

Let CP meet the circumcircle of triangle ABC at P', then <P'BP=<P'BA+<PBA=<PCA+<PBA=<PBC+<PCB=<P'PB. Thus P'P=P'B. Since <P'=<A, so we have $\angle{BPC}=90^{o}+\frac{A}{2}=\angle{BIC}$ . Thus B, P, I, C lie on the same circle (Let's say with centre $D$ ). But it is a well-known result that $D$ lies on line $AI$ , since we have $\angle{DIP}\leq 90^{o}$ , so $\angle{AIP}\geq 90^{o}$ . Hence $AP \geq AI$ with equality holds iff $I=P$ .

Attachments:

This post has been edited 2 times. Last edited by shobber, Jul 12, 2006, 12:36 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Yimin Ge

253 posts

Yimin Ge

#3 h Jul 12, 2006, 12:30 PM • 4 Y

Y by Adventure10, mathleticguyyy, jhu08, miloss.delfosontop

(Almost) same solution as mine.
This was very easy.....
Seems as tough this year will be a year with many many honorable mentions.....

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Ashegh

858 posts

Ashegh

#4 h Jul 12, 2006, 12:33 PM • 5 Y

Y by anser, Adventure10, jhu08, AylyGayypow009, miloss.delfosontop

Ok………………………………………………………………….

We have: $\angle PBA + \angle PCA = \angle PBC + \angle PCB$

If we add $\angle PBC + \angle PCB$ , to both side of the equality we have:

$2(\angle PBC + \angle PCB) = \angle PBA + \angle PCA + \angle PBC + \angle PCB = \angle B + \angle C$

then, it means that $\angle PBC + \angle PCB$ , is fixed for each $P$ .

then we conclude that $\angle BPC$ is also fixed, and it is clear that, the angle is equal to $\angle BIC$

then the locus of $P$ is arc $BIC$ .

We know that this arc is tangent to $AB,AC$ at $B,C$ respectively.

And also know that $AI$ goes through the center of circle $BIC$ .

Then $I$ is the nearest point of circle to $A$ .

And for each point $P$ on the circle we have:

$AI < AP$

and if $P$ is $I$ , it is clear that $AP = AI$ .

How ever it was nice and simple.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

perfect_radio

2607 posts

perfect_radio

#5 h Jul 12, 2006, 12:35 PM • 3 Y

Y by NewAlbionAcademy, Adventure10, jhu08

[spam]Finally a geo problem from IMO which I can solve.

It is a very nice restate of the fact that the intersection $J$ of the perpendicular bisector of $BC$ with $AI$ satisfies $JB=JI=JC$ .[/spam]

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Arne

3660 posts

Arne

#6 h Jul 12, 2006, 12:55 PM • 4 Y

Y by Adventure10, jhu08, Mango247, ehuseyinyigit

You don't need to construct the point P', Shobber, but of course that works.

I like the idea of having an easy question for problem 1, but isn't this too simple?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

vedran6

122 posts

vedran6

#7 h Jul 12, 2006, 1:08 PM • 4 Y

Y by Adventure10, jhu08, Mango247, and 1 other user

just prove that BIPC is cyclic than let intersection of AI and circle aroun BIPC be V and AP and that circle be W.
Than using angles we can prove that IBV is 90 from which we conclude IV is diameter.(2 radius).Than IV>=PW and we use potention of point A,we have AI*AV=AP*AW and if we assume that AI>AP we easy get a contradiction.From that we conclude AP>=AI and easy solve the cases of equality

I hope this is enough for 7

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

mahbub

912 posts

mahbub

#8 h Jul 12, 2006, 3:14 PM • 3 Y

Y by Adventure10, jhu08, Mango247

The problem is too much easy at IMO stage.

My solution is:

Actually P is a point on the circum circle of BIC and of course inside ABC. Let P is on arc BI.

To prove, $AP \geq AI$ it is sufficient to prove $\angle{AIP}\geq \angle{API}$ .

$\angle{AIP}=360-\angle{AIC}-\angle{PIC}=360-(90+B/2)-(180-\angle{PBC})=90-B/2+\angle{PBC}$

$\angle{API}=360-\angle{APB}-\angle{BPI}=360-\angle{APB}-(180-\angle{BCI})=180-\angle{APB}+C/2$

$\angle{AIP}\geq \angle{API}$
implies, $90-B/2+\angle{PBC}\geq 180-\angle{APB}+C/2$
implies, $90-B/2+\angle{PBI}+\angle{IBC}\geq 180-\angle{APB}+C/2$
implies, $\angle{PBI}+\angle{APB}\geq 90+C/2$
implies, $\angle{PBI}+\angle{APB}\geq \angle{AIB}$

But, $\angle{APB}\geq \angle{AIB}$ Since P is inside AIB.

So the above inequality is true. Equality when P is I and then $\angle{PBI}=0$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

shobber

3498 posts

shobber

#9 h Jul 12, 2006, 3:54 PM • 3 Y

Y by Adventure10, jhu08, Mango247

What happened to my attached picture? It was there when posted the proof but now it is gone!

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Benz

11 posts

Benz

#10 h Jul 12, 2006, 4:57 PM • 3 Y

Y by Adventure10, jhu08, Mango247

As I read "triangle triviality" I thought that it just can't happen at the IMO;
but in a while I realized it was true - the only facts I used to solve this problem are very, very basic things like sum of angles in a triangle

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

abdurashidjon

119 posts

abdurashidjon

#11 h Jul 12, 2006, 5:02 PM • 3 Y

Y by Adventure10, jhu08, Mango247

My solution was as shobbers solution but at the end i have used inequalities defferently. I hope my solutin is also true

Abdurashid

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Joel_Larsson

2 posts

Joel_Larsson

#12 h Jul 12, 2006, 5:54 PM • 2 Y

Y by Adventure10, jhu08

I had a much longer solution to this, not sure if that's a good thing though.. probably isn't. Most people here seem to have got this one right, so the limits for medals are likely to be higher this year than last.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

bodom

123 posts

bodom

#13 h Jul 12, 2006, 6:20 PM • 4 Y

Y by Adventure10, jhu08, SPHS1234, Mango247

i couldn't believe that this is an imo problem when i saw it.it only took me 10 minutes to solve it(and i'm not good at geometry).i imagine all of you guys did it in....3 min

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

jmerry

12096 posts

jmerry

#14 h Jul 12, 2006, 7:43 PM • 4 Y

Y by Adventure10, jhu08, Mango247, and 1 other user

It was a $\le$ 10 minute problem for me, too.

My take on the problem:

First, we note that $P$ must be a point on circle $BIC$ ; $\angle PBA+\angle PCA-\angle PBC-\angle PCB=\angle CBA+\angle BCA-2\angle PBC-2\angle PCB$
$=(180^\circ-\angle BAC)-2(180^\circ-\angle BPC)=2\angle BPC-\angle BAC-180^\circ$ . This is zero if $P=I$ (by the original form), so $\angle BPC=\angle BIC$ and $BPIC$ is cyclic.

It took a lot less time to notice the fact than to write this out; anyone that has done olympiad angle chasing should see the cyclic quadrilateral immediately.

Now, we want to show that $I$ is the closest point to $A$ on circle $BIC$ . Equivalently, the line $AI$ passes through the center of circle $BIC$ . The center must lie on the perpendicular bisector of $BC$ , which meets the bisector $AI$ of angle $BAC$ at $A'$ on the circumcircle of $ABC$ (the midpoint of arc $BC$ ). We want to show that $A'$ is the center of circle $BIC$ .
Chasing some angles, $\angle IA'C=\angle AA'C=\angle ABC=2\angle IBC$ . Similarly, $\angle IA'B=2\angle ICB$ . The only point other than $I$ making these (signed) angles is the center of circle $BIC$ , so $A'$ is that center and we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Amir.S

786 posts

Amir.S

#15 h Jul 12, 2006, 8:03 PM • 9 Y

Y by wiseman, Starlex, Kriket, The_Outsider, raven_, Adventure10, jhu08, rayfish, Mango247

ooooooooooooops! I can't beleive it was IMO problem,really easy.(It took me 30 seconds to solve)

we have $\angle ABP+\angle PCA=\angle PBC+\angle PCB=90^{\circ}-\frac{A}{2}\rightarrow \angle BPC=90^{\circ}+\frac{A}{2}$
hence points $B,P,I,C$ are on a circle,we know that $AI$ pass through circumcenter of $BIC$ therefore $AI\le AP$

This post has been edited 1 time. Last edited by Amir.S, Jul 13, 2006, 7:40 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

KHOMNYO2

98 posts

KHOMNYO2

#137 h Jul 11, 2024, 3:03 AM

Y by

$BPIC$ is cyclic then the rest is trivial.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ezpotd

1365 posts

ezpotd

#138 h Aug 27, 2024, 10:56 PM

Y by

Clearly $BPIC$ is cyclic, so the unique closest point on the circle to $A$ is the intersection of segment $AM_A$ and the circle (since $M_A$ is the center), where $M_A$ is the arc midpoint, but this is just $I$ , so we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Mathandski

777 posts

Mathandski

#139 h Aug 28, 2024, 4:08 AM

Y by

Subjective Rating (MOHs) $\text{[math]}$

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

pie854

253 posts

pie854

#140 h Sep 4, 2024, 8:33 PM

Y by

Solved with some help... my first IMO geo btw

We have $\begin{align*}\angle PBC+\angle PCB = & \ \angle PBA+\angle PCA \\ \angle ABC+\angle ACB = & \ 2 \left (\angle PBA+\angle PCA\right) \\ \frac 12\left (180^\circ-\angle BAC\right) = & \ 180^\circ -\angle PAB-\angle APB+180^\circ-\angle PAC-\angle APC \\ 90^\circ-\frac 12 \angle BAC = & \ 360^\circ-\left(\angle APB+\angle APC\right )-\angle BAC \\ 90^\circ+\frac 12 \angle BAC= & \ \angle BPC \\ \angle BIC = & \ \angle BPC.\end{align*}$ Now let $I_A$ be the $A$ -excenter of $\triangle ABC$ . Then $BICI_A$ is cyclic, by the Incenter Excenter Lemma. Let the center of $(BICI_A)$ be $O$ . Since $180=\angle BI_AC+\angle BIC=\angle BI_AC+\angle BPC,$ it follows that $BPCI_A$ is cyclic as well. So, $P$ must lie on $(BCI_A)$ . So by the triangle inequality we have $OA \leq AP+PO \implies OI+IA \leq AP+OI \implies IA \leq AP,$ and the equality obviously holds if and only if $I=P$ , as desired.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9.36cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -1.41, xmax = 1.71, ymin = -1.04, ymax = 0.7; /* image dimensions */ pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); /* draw figures */ draw(circle((0.,0.), 0.5), linewidth(0.8) + yellow); draw((-0.2995682831586557,0.4003234239029431)--(-0.4,-0.3), linewidth(0.8)); draw((-0.2995682831586557,0.4003234239029431)--(0.4,-0.3), linewidth(0.8)); draw((-0.4,-0.3)--(0.4,-0.3), linewidth(0.8)); draw(circle((0.,-0.5), 0.447213595499958), linewidth(0.8) + red); draw((-0.4,-0.3)--(-0.14119247578894345,-0.07565970639050944), linewidth(0.8) + green); draw((-0.4,-0.3)--(0.14119247578894334,-0.9243402936094904), linewidth(0.8) + green); draw((0.14119247578894334,-0.9243402936094904)--(0.4,-0.3), linewidth(0.8) + green); draw((-0.14119247578894345,-0.07565970639050944)--(0.4,-0.3), linewidth(0.8) + green); draw((-0.2995682831586557,0.4003234239029431)--(0.14119247578894334,-0.9243402936094904), linewidth(0.8)); draw((-0.2995682831586557,0.4003234239029431)--(0.,-0.05278640450004202), linewidth(0.8) + blue); /* dots and labels */ dot((-0.2995682831586557,0.4003234239029431),linewidth(4.pt)); label("$A$", (-0.2838752322507354,0.43122566059848144), NE * labelscalefactor); dot((-0.4,-0.3),linewidth(4.pt)); label("$B$", (-0.384104276037203,-0.26636848415533076), NE * labelscalefactor); dot((0.4,-0.3),linewidth(4.pt)); label("$C$", (0.4177280742545375,-0.26636848415533076), NE * labelscalefactor); dot((-0.14119247578894345,-0.07565970639050944),linewidth(4.pt) + uuuuuu); label("$I$", (-0.12350876219238732,-0.04185542607364408), NE * labelscalefactor,uuuuuu); dot((0.,-0.5),linewidth(4.pt) + uuuuuu); label("$O$", (0.016811899108667254,-0.46682657172826536), NE * labelscalefactor,uuuuuu); dot((0.14119247578894334,-0.9243402936094904),linewidth(4.pt) + uuuuuu); label("$I_A$", (0.15713256040972184,-0.8917977173828866), NE * labelscalefactor,uuuuuu); dot((0.,-0.05278640450004202),linewidth(4.pt) + uuuuuu); label("$P$", (0.016811899108667254,-0.021809617316350626), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

This post has been edited 4 times. Last edited by pie854, Sep 7, 2024, 3:40 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Amkan2022

2056 posts

Amkan2022

#141 h Sep 9, 2024, 2:57 AM

Y by

outline for storage

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

lnzhonglp

120 posts

lnzhonglp

#142 h Nov 26, 2024, 3:55 AM

Y by

Let $M$ be the midpoint of arc $\widehat{BC}$ of $(ABC)$ . The conditions imply $\angle BPC = 90^\circ + \frac{\angle{A}}{2},$ implying that $P$ lies on the circle centered at $M$ passing through $B$ and $C$ . Then by the triangle inequality, $AP + PM \geq AM \implies AP \geq AI,$ with equality when $P = I$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Polkishtrn_Kazakhstan

5 posts

Polkishtrn_Kazakhstan

#143 h Nov 27, 2024, 9:26 AM

Y by

We now that BPIC-cyclic by angle chasing.And Circle with center A an radius AI is tangent to (BIC) (easy by tangent from I to (BIC))And AP>AI when P!=I.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Maximilian113

592 posts

Maximilian113

#144 h Jan 2, 2025, 1:50 AM

Y by

Observe that $\angle PBA + \angle PCA +\angle PBC + \angle PCB = 180^\circ - \angle A \implies \angle PBA + \angle PCA = 90^\circ - \frac12 \angle A \implies \angle BPC = 90^\circ + \frac12 \angle A.$ But it is well-known that this is equal to $\angle BIC,$ therefore $B, P, I, C$ are concyclic with center $O.$ But by the Incenter-Excenter Lemma, $AI$ passes through $O,$ therefore $AP \geq AI$ is obvious, with equality iff $P=I.$ QED

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Markas

150 posts

Markas

#145 h May 11, 2025, 6:20 PM

Y by

Let $\angle ABP = x$ , $\angle ACP = y$ , $\angle ABC = \beta$ , $\angle ACB = \gamma$ . From the condition $\angle PBA + \angle PCA = \angle PBC + \angle PCB$ , we have that $x + (\frac{\gamma}{2} + y) = (\beta - x) + (\frac{\gamma}{2} - y)$ $\Rightarrow$ $2(x + y) = \beta$ $\Rightarrow$ $\angle BPC = \angle ABP + \angle BAC + \angle ACP = \alpha + x + y + \frac{\gamma}{2} = \alpha + \frac{\beta}{2} + \frac{\gamma}{2} = 90 + \frac{\alpha}{2} = \angle BIC$ $\Rightarrow$ $\angle BPC = \angle BIC = 90 + \frac{\alpha}{2}$ $\Rightarrow$ $P \in (BIC)$ . Now $\angle AIP = 360 - \angle AIC - \angle PIC = 360 - (90 + \frac{\beta}{2}) - (180 - \beta + x) = 90 + \frac{\beta}{2} - x$ . We have that $\angle ABI = \angle ABP + \angle PBI$ $\Rightarrow$ $\frac{\beta}{2} \geq x$ $\Rightarrow$ $\angle AIP \geq 90^{\circ}$ $\Rightarrow$ $AP \geq AI$ , where equality is obtained when $P \equiv I$ . We are ready.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

mathnerd_101

1518 posts

mathnerd_101

#146 h Jun 6, 2025, 8:28 PM • 1 Y

Y by peace09

Note that adding $\angle{PBC} + \angle{PCB}$ to both sides of our given equation gives us $2(\angle{PBC} + \angle{PCB}) = \angle{PBA} + \angle PCA + \angle PBC + \angle PCB = \angle B + \angle C.$ Furthermore, we know that $\angle BPC = 180 - (\angle PCB + \angle PBC) = 180 - \frac{\angle ABC + \angle ACB}{2}$ by what we just found. Thus, $\angle BPC = 90 + \frac{\angle A}{2} = \angle BIC.$ Thus, we know that $P$ lies on the circle containing $B,I,C$ . By the incenter-excenter lemma, we know that $A,I,$ and the center of circle $(BPC)$ are collinear, so the shortest distance between $AP$ and $AI$ of $AP=AI$ holds if $P=I,$ and all other scenarios result in $AP > AI,$ as desired.

Remark: Xiooix I'm back on my otis grind or something like that
Remark 2: Xoinky sploinky 1500th post

This post has been edited 2 times. Last edited by mathnerd_101, Jun 6, 2025, 8:29 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

mudkip42

433 posts

mudkip42

#147 h Jun 16, 2025, 11:56 PM

Y by

Note that $\angle PBA + \angle PCA = 360^{\circ}-A-(360^{\circ}-\angle BPC)=\angle BPC-A$ and $\angle PBC+\angle PCB=180^{\circ}-\angle BPC$ where $A$ represents $\angle BAC$ . Thus, $\angle BPC = 90^{\circ}+\frac{A}{2}=\angle BIC,$ so $BIPC$ is cyclic. Extend $AI$ to meet $(ABC)$ at $L.$ Then $L$ is the circumcenter of $BIPC$ by Incenter/Excenter Lemma, and since $A, I, L$ are collinear, $AI$ is the shortest distance from $A$ to $(BPIC).$ Thus, $AP \ge AI$ with equality occuring iff $P=I,$ so we're done. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

cappucher

101 posts

cappucher

#148 h Jun 23, 2025, 10:05 PM

Y by

Angle chase to find that $\angle{A} = 180^{\circ} - 2(\angle{PBA} + \angle{PCA})$ . Recall that $\angle{BIC} = 90^{\circ} + \frac{1}{2}\angle{A}$ . Thus,

$\angle{BIC} = 90^{\circ} + 90^{\circ} - \angle{PBA} - \angle{PCA} = \angle{BPC}$
Since $\angle{BIC} = \angle{BPC}$ , quadrilateral $BPIC$ is cyclic.

Now, apply the incenter-excenter lemma by extending $AI$ to meet $(ABC)$ at $L$ . From this lemma and by drawing the circle with center $L$ and radius $LI$ , we find that $B$ , $P$ , $I$ , and $C$ all lie on this circle. Finally, we apply the triangle inequality:

$LP + PA \geq LA$
We can expand $LA$ as $LI + IA$ . Because $P$ and $I$ both lie on the circle and thus are both equal to the radius of the circle, we can cancel out these terms and finally conclude that

$AP \geq AI$
with the equality case occurring only when $\triangle{APL}$ is a degenerate triangle; i.e., when $P = I$ .

This post has been edited 2 times. Last edited by cappucher, Jun 23, 2025, 10:06 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Ilikeminecraft

747 posts

Ilikeminecraft

#149 h Jul 4, 2025, 6:02 AM

Y by

Solved with a_smart_alecks buh buh buh skibidi skuh

Trivially note that $BCIP$ is cyclic due to the angle chase. Use Chicken foot to conclude.

This post has been edited 1 time. Last edited by Ilikeminecraft, Jul 4, 2025, 6:02 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Kempu33334

852 posts

Kempu33334

#150 h Jul 8, 2025, 10:33 PM

Y by

Diagram

We can rewrite the condition as $\angle PBA+\angle PCA = \angle B - \angle PBA + \angle C - \angle PCA$ which simplifies to $\angle PBA + \angle PCA = \frac{\angle B}{2}+\frac{\angle C}{2}.$ We can rearrange this to $\angle PBA - \frac{\angle B}{2} = \frac{\angle C}{2}-\angle PCA.$ This implies that $P$ must lie on $(BIC)$ .

It is well known that the minimum distance from any point to a circle can be found by finding the distance from that point to intersection of the segment with endpoints of the center of the circle and that point. In addition, by the Incenter-Excenter Lemma, we know that $A$ , $I$ and the center of $(BIC)$ are collinear, so the minimum distance from $A$ to $(BIC)$ is $AI$ . Hence, we must have $AP \ge AI$ for any $P$ , with equality that holds if and only if $P = I$ , as required. $\square$