APMO 2012 #1

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

APMO 2012 #1

G H J

Reveal topic tags

inequalities proposed

L

G H BBookmark kLocked kLocked NReply

Source: APMO 2012 #1

The post below has been deleted. Click to close.

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

202 posts

#1 h Apr 2, 2012, 3:15 PM • 5 Y

Y by ahmedosama, jishu2003, Adventure10, Mango247, and 1 other user

Let $P$ be a point in the interior of a triangle $ABC$ , and let $D, E, F$ be the point of intersection of the line $AP$ and the side $BC$ of the triangle, of the line $BP$ and the side $CA$ , and of the line $CP$ and the side $AB$ , respectively. Prove that the area of the triangle $ABC$ must be $6$ if the area of each of the triangles $PFA, PDB$ and $PEC$ is $1$ .

This post has been edited 1 time. Last edited by syk0526, Apr 4, 2012, 6:43 AM

Z K Y

The post below has been deleted. Click to close.

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

2903 posts

#2 h Apr 2, 2012, 6:06 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

I'm not seeing how this is an inequality problem, it should be moved to the geometry forums

A very heavyily algebraic solution:

Let $a = [PDC], b = [PAE], c = [BFP]$ . Then:
$1/a = (2 + c)/(1 + b + a) \implies 1 + b = a + ac$
$1/c = (2+b)/(1+c+a) \implies 1 + a = c + bc$
$1/b = (2+a)/(1+b+c) \implies 1 + c = b + ab$
$a = \frac{b+1}{c+1}$
$b = \frac{c+1}{a+1}$
$c = \frac{a+1}{b+1}$

$b = \frac{c+1}{\frac{b+1}{c+1}+1} = \frac{(c+1)^2}{b+c+2}$
$b^2 + b(2+c) - (c+1)^2 = 0$
$b = \frac{-2-c \pm \sqrt{(c+2)^2 + 4(c+1)^2}}{2} = \frac{-2-c + \sqrt{5c^2 + 12c + 8}}{2}$
$b+1 = \frac{-c + \sqrt{5c^2 + 12c + 8}}{2}$
Hence $a = \frac{-c + \sqrt{5c^2 + 12c + 8}}{2c + 2}$
$a+1 = \frac{c+2 + \sqrt{5c^2 + 12c + 8}}{2c + 2}$

$c = \frac{\frac{c+2 + \sqrt{5c^2 + 12c + 8}}{2c + 2}}{\frac{-c + \sqrt{5c^2 + 12c + 8}}{2}}$
$c^2 + c = \frac{c+2 + \sqrt{5c^2 + 12c + 8}}{-c + \sqrt{5c^2 + 12c + 8}}$
$c^2 + c = \frac{(c+2 + \sqrt{5c^2 + 12c + 8})(c + \sqrt{5c^2 + 12c + 8})}{4c^2 + 12c + 8}$
$4c^4 + 16c^3 + 20c^2 + 8c = 6c^2 + 14c + 8 + (2c+2)\sqrt{5c^2 + 12c + 8}$
$4c^4 + 16c^3 + 14c^2 - 6c - 8 = (2c+2)\sqrt{5c^2 + 12c + 8}$
$2c^3 + 6c^2 + c - 4 = \sqrt{5c^2 + 12c + 8}$
$(2c^3 + 6c^2 + c - 4)^2 - 5c^2 - 12c - 8 = 0$
$4(c-1)(c+1)(c+2)(c^3 + 4c^2 - 3c - 1) = 0$
Observe that $c=1$ is the only positive root except a root around $0.25$ of the cubic. You can bound the root between $0.2$ and $0.25$ . But for that root $2c^3 + 6c^2 + c - 4 < 0$ and hence it is not actually a solution for $c$ . Hence the only solution is $c=1$ , so $a=b=1$ as well and therefore $[ABC] = 6$ .

Z K Y

The post below has been deleted. Click to close.

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

958 posts

#3 h Apr 2, 2012, 6:29 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

No need to do all that, lol.. Just use Ceva's Theorem to get $abc = 1.$ , and then it's a simple application of basic inequalities with the equations you found..

Z K Y

The post below has been deleted. Click to close.

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

2903 posts

#4 h Apr 2, 2012, 6:35 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Yes that's true, the solution I had above was very dumb but straightforward, it only took about 10~15 minutes to bash out actually.

Z K Y

The post below has been deleted. Click to close.

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

958 posts

#5 h Apr 2, 2012, 6:45 PM • 4 Y

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

Hmm actually you don't need inequalities.. Summing up your equations you get $ab + bc +ca = 3$ , and with $abc = 1$ , we have :
$b+b^2 = ab + 1$ $c+ c^2 = bc+1$ $a +a^2 = ac + 1$
Summing again, we get $a+b+c + (a+b+c)^2 - 12 = 0 \implies a+b+c=3.$

How did you factor that huge polynomial and get the root estimations of your cubic so quickly without a calculator?

Z K Y

The post below has been deleted. Click to close.

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

2903 posts

#6 h Apr 2, 2012, 8:04 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Well $c=1$ is obviously a root. Playing around a little I found $-1,-2$ are also roots so I factored it then, so it sufficed to consider the cubic term. The cubic has three real roots (easy to verify) and there is only one positive root which happens to be between 0 and 1. Then guessing I found the root was less than $0.5$ , then less than $0.25$ which was enough to establish the root thing was negative yielding $c=1$ as the only actual solution.

Z K Y

The post below has been deleted. Click to close.

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

555 posts

#7 h Apr 2, 2012, 8:39 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

From $ab+bc+ac = 3$ and $abc = 1$ you can say $a=b=c=1$ because of $AM - GM$ .. I think this was a good problem 1, not completely straightforward but very natural if you know how to prove ceva for example.

Z K Y

The post below has been deleted. Click to close.

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

417 posts

zhaoli

#8 h Apr 3, 2012, 1:04 PM • 4 Y

Y by Adventure10 and 3 other users

See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=55209.

It is posted in mathlinks more than 6 years ago.

Z K Y

The post below has been deleted. Click to close.

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

3 posts

SRKTK

#10 h Feb 27, 2019, 6:01 PM • 2 Y

Y by Adventure10, Mango247

how do you get ab+bc+ca=3 ??

Z K Y

The post below has been deleted. Click to close.

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

Synthetic_Potato

114 posts

Synthetic_Potato

#11 h Feb 27, 2019, 6:18 PM • 4 Y

Y by AlanLG, Adventure10, Mango247, Mango247

Here is a more general problem.

$P$ is a point inside triangle $ABC$ , and $AP, BP,CP$ meet the opposite sides at $D,E,F$ respectively. If the areas of three of the six triangles so formed are equal then $P$ is the centroid of triangle $ABC$ .

Z K Y

The post below has been deleted. Click to close.

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

1107 posts

#12 h Feb 27, 2019, 7:37 PM • 1 Y

Y by Adventure10

APMO 2012 P1 wrote:

Let $P$ be a point in the interior of a triangle $ABC$ , and let $D, E, F$ be the point of intersection of the line $AP$ and the side $BC$ of the triangle, of the line $BP$ and the side $CA$ , and of the line $CP$ and the side $AB$ , respectively. Prove that the area of the triangle $ABC$ must be $6$ if the area of each of the triangles $PFA, PDB$ and $PEC$ is $1$ .

Maybe Bary :maybe:

:maybe:

Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ ,let $P=(p,q,r)$ Its easy to see that $D=\left(0,\dfrac{q}{q+r},\dfrac{r}{q+r}\right),E=\left(\dfrac{p}{p+r},0,\dfrac{r}{r+p}\right),F=\left(\dfrac{p}{p+q},\dfrac{q}{p+q},0\right)$ .The area of the respective triangles are - $X=\left\{\Delta{PAE}=\dfrac{qr}{r+p},\Delta{PEC}=\dfrac{pq}{r+p}\right\},Y=\left\{\Delta{PCD}=\dfrac{pq}{q+r},\Delta{PDB}=\dfrac{rp}{q+r}\right\},Z=\left\{\Delta{PBF}=\dfrac{rp}{p+q},\Delta{PFA}=\dfrac{qr}{p+q}\right\}$
Observe that the area of any group $\in \{p,q,r\}$ .Now of any 3 selected triangles 2 must belong to one of the group $\left(X,Y,Z\right)$ .So we have to solve the equation $\dfrac{pq}{q+r}=\dfrac{qr}{r+p}=\dfrac{rp}{p+q}$ .I feel sleepy so I will continue tommorow anyone wanting to modify my work is welcomed.

$\mathbf{NOTE}$ -I originally had created the sets $(X,Y,Z)$ to solve the problem proposed by @synthetic_potato but.....

This post has been edited 4 times. Last edited by Pluto1708, Feb 27, 2019, 7:39 PM
Reason: j

Z K Y

The post below has been deleted. Click to close.

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

1594 posts

#13 h Mar 3, 2020, 9:34 AM • 1 Y

Y by mijail

Quite hard for a #1 (especially if you are anti-bash)

First, we make some synthetic observation. Since $[PFA]=[PDB]$ , we have
$\frac{AP}{PD} = \frac{\mathrm{dist}(B,AD)}{\mathrm{dist}(F,AD)} = \frac{AB}{AF}$ and cyclic relation holds. Now we resort to barycentric coordinates w.r.t. $\triangle ABC$ . Set $P=(a:b:c)$ . Then we clearly have $AP : PD = b+c:a$ and $AB : AF = a+b : b$ . Thus
$\frac{b+c}{a} = \frac{a+b}{b}\implies a(a+b) = b(b+c).$ Cyclic relations hold so $a(a+b)=b(b+c)=c(c+a)$ .

To solve this system of equations, assume that $a>b$ , then $a+b<b+c$ $\implies a<c$ , which in turn implies $a+b>c+a$ $\implies b>c$ , contradiction. Similarly $a<b$ causes contradiction hence $a=b=c$ . This means $P$ is a centroid and the result became obvious.

Z K Y

The post below has been deleted. Click to close.

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

572 posts

anser

#14 h Sep 14, 2020, 4:31 AM

Y by

Let $[PCD] = x, [PAE] = y, [PBF] = z$ . By Ceva's theorem, $xyz = 1$ . Also $\frac{CP}{PF} = \frac{x+1}{z} = y+1$ , so $x+1 = yz + z$ (similarly, $y+1 = zx + x$ and $z+1 = xy+y$ ). If $x = 1$ , then we can solve to get $x=y=z=1$ . Otherwise, WLOG $x > 1$ and $y, z < 1$ , or $x < 1$ and $y, z > 1$ . In the first case, $2 < x+1 = yz+z < 2$ , and in the second case, $2 > x + 1 = yz + z > 2$ , contradiction.

Z K Y

The post below has been deleted. Click to close.

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

1602 posts

#15 h Sep 18, 2021, 8:25 AM

Y by

2 liner Hype

This post has been edited 1 time. Last edited by starchan, Sep 18, 2021, 8:26 AM

Z K Y

The post below has been deleted. Click to close.

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

Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#16 h Feb 5, 2022, 6:19 AM

Y by

Really Nice problem
Let [PFB] = x, [PDC] = y and [PEA] = z.
Claim1 : xyz = 1.
Proof : By Ceva Theorem we have 1/x . 1/y . 1/z = 1 so xyz = 1.

Claim2 : xy + yz + zx = 3.
Proof : Note that [APB]/[CPB] = [AEP]/[CEP] so x+1/y+1 = z so x+1 = zy + z with same approach we have y+1 = xz + x and z+1 = yx + y so xy + yz + zx = 3.
Now with AM - GM we have xy = yz = zx so x = y = z = 1 so [ABC] = 6.
we're Done.

Z K Y

The post below has been deleted. Click to close.

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

67 posts

ZNatox

#17 h Jan 13, 2023, 1:41 PM • 1 Y

Y by Mango247

Let $x=[\triangle APE], y=[\triangle BPF], z=[\triangle CPD]$ . By Ceva: $xyz=1$ . Also $\frac{[\triangle CPE]}{[\triangle CPB]}=\frac{[\triangle APE]}{[\triangle APB]}$ , or $1+y=x+xz=x+\frac1y$ or $(y+1)\cdot \frac{y-1}y=x-1$ . If one of $x,y,z$ is one, then the others are clearly one. Otherwise, by multiplying cyclically the last equation, we get $(x+1)(y+1)(z+1)=1$ , which is impossible.

This post has been edited 2 times. Last edited by ZNatox, Jan 13, 2023, 1:41 PM

Z K Y

The post below has been deleted. Click to close.

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

4176 posts

#18 h Jul 5, 2023, 4:28 AM

Y by

Lemma: If $x,y,z$ are positive real numbers such that $xyz=1, xy+xz+yz=3,$ then $x=y=z=1.$ Divide the second equation by $xyz$ so that $1/x+1/y+1/z=3.$ Let $a=1/x,b=1/y,c=1/z.$ Then, this becomes $a+b+c=3,abc=1.$ By AM-GM, $3=a+b+c\geq 3\sqrt[3]{abc}=3,$ so we must have $a=b=c$ and thus $x=y=z=1.$

Let $[APE]=x,[BPF]=y,[CPD]=z.$ Then, $xyz=1$ by the fact that $x=AE/CE$ etc. and Ceva. Furthermore, $\frac{[ABD]}{[ACD]}=\frac{BD}{CD}=\frac{[BPD]}{[CPD]}.$ Hence, $\frac{y+2}{x+z+1}=\frac{1}{z}.$ This rearranges to $yz=x-z+1.$ Similarly, $zx=y-x+1,xy=z-y+1.$ Adding these three equations, we get that $xy+xz+yz=3.$ By our lemma, it follows that $x=y=z=1$ , 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.

962 posts

#20 h Mar 3, 2024, 8:56 AM

Y by

Denote $[BPF] = a, [PDC] = b, [PEA] = c$ .
Observe from Ceva's theorem that $abc = 1$ .
Observe that $b = \frac{CD}{DB} = \frac{b+c+1}{a+2} \implies b + c + 1 = ba + 2b.$ Similarly, $a+b+1 = ac + 2a$ and $c + a + 1 = cb + 2c.$ Adding up, we see that $ab + bc + ca = 3.$ But now $\frac{ab + bc + ca}{3} = \frac{3}{3} = 1 = 1^{\frac{2}{3}} = (abc) ^{\frac{2}{3}}$ , and thus the equality case of AM-GM holds. Therefore $ab = bc = ca = 1 \implies a = b = c = 1$ . Therefore the area of $ABC$ is $a + b + c + 3 = 6$ . $\square$

Z K Y

The post below has been deleted. Click to close.

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

1455 posts

#21 h Mar 4, 2024, 5:10 AM

Y by

We proceed by barycentric coordinates. Let $P = (a, b, c)$ . Then we have cyclically that $\frac{[AFP]}{[ABC]} = \frac{bc}{a+b}$ .

Thus cyclically, we have $\frac{bc}{a+b} = \frac{ca}{b+c}$ so $b(b+c) = a(a+b)$ so $b - ab = a - ac$ so $b = a(1+b-c)$ .

Now multiplying the final equation cyclically, we have $\prod_{\textrm{cyc}}(1+b-c) = 1$ , so by AM-GM it follows that $a = b = c$ . Thus $P$ is the centroid, so it follows immediately that $[ABC] = 6$ .

Z K Y

The post below has been deleted. Click to close.

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

154 posts

#22 h May 11, 2024, 9:58 AM

Y by

Solution

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a