Inequality with convex pentagons

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

Inequality with convex pentagons

G H J

Reveal topic tags

geometry unsolved

L

G H BBookmark kLocked kLocked NReply

Source: Brazil MO #3

The post below has been deleted. Click to close.

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

958 posts

#1 h Oct 20, 2011, 12:44 AM • 2 Y

Y by Davi-8191, Adventure10

Prove that, for all convex pentagons $P_1 P_2 P_3 P_4 P_5$ with area 1, there are indices $i$ and $j$ (assume $P_7 = P_2$ and $P_6 = P_1$ ) such that:

$\text{Area of} \ \triangle P_i P_{i+1} P_{i+2} \le \frac{5 - \sqrt 5}{10} \le \text{Area of} \ \triangle P_j P_{j+1} P_{j+2}$

Z K Y

The post below has been deleted. Click to close.

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

4302 posts

#2 h Oct 20, 2011, 11:32 PM • 1 Y

Y by Adventure10

All indices are taken $\pmod{5}$ here.

Case 1: The triangle of maximal area is $\triangle{P_{i-1}P_iP_{i+1}}$ for some $i$ . Then convex quadrilateral $P_{i+1}P_{i+2}P_{i-2}P_{i-1}\subseteq \triangle{P_{i+1}P'_iP_{i-1}}$ , where $P_iP_{i+1}P'_iP_{i-1}$ is a parallelogram, so $[P_{i-1}P_iP_{i+1}]\ge 1/2$ while $[P_{i-2}P_{i+1}P_{i+2}]\le [P_{i-1}P_{i+1}P_{i+2}]$ and $[P_{i+2}P_{i-1}P_{i-2}]\le [P_{i+1}P_{i-1}P_{i-2}]$ (easy to see by considering the intersection of rays $P_{i-1}P_{i-2}$ and $P_{i+1}P_{i+2}$ ) force
$\begin{align*} \min([P_{i-1}P_{i-2}P_{i+2}],[P_{i+1}P_{i+2}P_{i-2}])\le\frac{[P_{i+1}P_{i+2}P_{i-2}P_{i-1}]}{2} &=\frac{1-[P_{i-1}P_iP_{i+1}]}{2} \\ &\le\frac{1}{4}<\frac{5-\sqrt{5}}{10}. \end{align*}$ (Clearly $1/2=5/10>(5-\sqrt{5})/10$ as well.)

Case 2: The triangle of maximal area is $\triangle{P_{i-2}P_iP_{i+2}}$ for some $i$ ; we note that $\phi(5-\sqrt{5})/10=\sqrt{5}/5$ where $\phi=(1+\sqrt{5})/2$ is the golden ratio.

Suppose that $[P_{j-1}P_jP_{j+1}]>(5-\sqrt{5})/10$ for all $j$ ; then $[P_{j-2}P_jP_{j+2}]<\sqrt{5}/5$ for all $j$ , contradicting 2002 G5 (by convexity, it's not hard to see that each "central triangle" has area exceeding that of at least one of its two bordering triangles, and so cannot be the minimal triangle).

On the other hand, if $[P_{j-1}P_jP_{j+1}]<(5-\sqrt{5})/10$ for all $j$ (so $[P_{j-2}P_jP_{j+2}]>\sqrt{5}/5$ for all $j$ ), then take an area-preserving affine transformation sending $\triangle{P_{i-2}P_iP_{i+2}}$ into a golden triangle (so $P_iP_{i\pm2}/P_{i-2}P_{i+2}=\phi$ , where $t(t\phi)\sin{72^\circ}/2=[P_{j-2}P_jP_{j+2}]>\sqrt{5}/5$ implies $P_{i-2}P_{i+2}=t>\sqrt{2\sqrt{5}/(5\phi\sin{72^\circ})}$ ) and construct $Q_{i-1}\in P_iP_{i-2}$ and $Q_{i+1}\in P_iP_{i+2}$ such that $P_{i\pm2}Q_{i\pm1}/P_{i\pm2}P_i=(5-\sqrt{5})/(5t^2\phi\sin{72^\circ})=s=1-r$ (where $r=P_iQ_{i\pm1}/P_{i\pm2}P_i$ ). Because $[Q_{i\pm1}P_{i-2}P_{i+2}]=(5-\sqrt{5})/10$ , we must have $P_{i-1},P_{i+1}$ in between $Q_{i-1}Q_{i+1}$ and $P_{i-2}P_{i+2}$ . Let $R_{i\pm1}=P_iP_{i\pm1}\cap Q_{i-1}Q_{i+1}$ (obviously $Q_{i-1}Q_{i+1}\subseteq R_{i-1}R_{i+1}$ ); then

\[\begin{align*}
\frac{5-\sqrt{5}}{10} > [P_{i-1}P_iP_{i+1}] > [R_{i-1}P_iR_{i+1}]
&= [Q_{i-1}P_iQ_{i+1}]+[Q_{i-1}P_iR_{i-1}]+[Q_{i+1}P_iR_{i+1}] \\
&= r^2[P_{i-2}P_iP_{i+2}]+r[P_iR_{i-1}P_{i-2}]+r[P_iR_{i+1}P_{i+2}] \\
&= (r^2-r)[P_{i-2}P_iP_{i+2}]+r[P_iP_{i+1}P_{i+2}P_{i-2}P_{i-1}] \\
&= (r^2-r)[P_{i-2}P_iP_{i+2}]+r \\
&= (r^2-r)\frac{t^2\phi\sin{72^\circ}}{2}+r\\
&= -rs\frac{t^2\phi\sin{72^\circ}}{2}+r = -r\frac{5-\sqrt{5}}{5t^2\phi\sin{72^\circ}}\frac{t^2\phi\sin{72^\circ}}{2}+r  = r\left(\frac{5+\sqrt{5}}{10}\right)
\end{align*},\]

whence
$\frac{1}{\phi^2}=\frac{5-\sqrt{5}}{5+\sqrt{5}} > r = 1-\frac{5-\sqrt{5}}{5t^2\phi\sin{72^\circ}} > 1-\frac{5-\sqrt{5}}{2\sqrt{5}}=1-\frac{1}{\phi}=\frac{1}{\phi^2},$ which is absurd.

Z K Y

The post below has been deleted. Click to close.

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

17 posts

#3 h Oct 21, 2011, 12:46 PM • 1 Y

Y by Adventure10

Let $a=[P_1P_2P_3],b=[P_2P_3P_4],c=[P_3P_4P_5],d=[P_4P_5P_1]$ and $e=[P_5P_1P_2]$ ,
We can prove that $a+b+c+d+e=ab+bc+cd+de+ea+1.$ Is this fact useful to solve this problem? Thanks.

Z K Y

The post below has been deleted. Click to close.

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

1044 posts

mszew

#4 h Oct 21, 2011, 4:54 PM • 2 Y

Y by Adventure10, Mango247

Can it be generalized for other regular poligons?

Z K Y

The post below has been deleted. Click to close.

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

165 posts

rsa365

#5 h Oct 22, 2011, 1:04 AM • 3 Y

Y by proglote, Adventure10, Mango247

Lemma 1:
If $A,B$ and $C$ are collinear and $P \notin AB$ , then $\frac{AB}{BC}=\frac{[ABP]}{[BCP]}$ .
To prove, just see that triangle $ABP$ and $BCP$ have the same altitude $h$ , so $\frac{[ABP]}{[BCP]}=\frac{AB\cdot h/2}{BC\cdot h/2}=\frac{AB}{BC}$
Lemma 2:
Let $A,B,C,D$ a quadrilateral. If $P$ is in segment $AB$ , $[CPD]\le max([CAD],[CBD])$ .
Its true because $CPD,CAD$ and $CBD$ have the same bases, and clearly the altitude of $CPD$ is not greater than both altitudes of $CAD$ and $CBD$ .

Let $\frac{5-\sqrt{5}}{10}=\alpha$ . Suppose that $[P_{j-1}P_{j}P_{j+1}]<\alpha$ for all $j$ and let $Q=P_{1}P_{3} \cap P_{2}P_{5}$ .
$[P_{2}P_{3}P_{4}]<\alpha$ and $[P_{3}P_{4}P_{5}]<\alpha$ . By lemma 2, $[P_{3}P_{4}Q]<\alpha$ .
$[P_{1}P_{2}P_{3}]<\alpha$ and $[P_{1}P_{5}P_{4}]<\alpha$ , so $[P_{1}P_{3}P_{4}]>1-2\alpha$ .
By lemma 1, $\frac{P_{3}Q}{P_{1}P_{3}}=\frac{[P_{3}P_{4}Q]}{[P_{1}P_{3}P_{4}]}<\frac{\alpha}{1-2\alpha}$ $(1)$ .
$[P_{3}P_{4}P_{5}]<\alpha$ , so $[P_{1}P_{2}P_{3}P_{5}]>1-\alpha$
By lemma 1, $\frac{P_{1}Q}{P_{1}P_{3}}=\frac{[P_{1}QP_{5}]}{[P_{1}P_{3}P_{5}]}=\frac{[P_{1}QP_{2}]}{[P_{1}P_{3}P_{2}]}=\frac{[P_{1}QP_{5}]+[P_{1}QP_{2}]}{[P_{1}P_{3}P_{5}]+[P_{1}P_{3}P_{2}]}=\frac{[P_{1}P_{2}P_{5}]}{[P_{1}P_{2}P_{3}P_{5}]}<\frac{\alpha}{1-\alpha}$ $(2)$
If we sum $(1)$ and $(2)$ :
$\frac{P_{3}Q+P_{1}Q}{P_{1}P_{3}}<\frac{\alpha}{1-2\alpha}+\frac{\alpha}{1-\alpha}$ , then $1<\frac{\alpha}{1-2\alpha}+\frac{\alpha}{1-\alpha}=1$ an absurd, so there is one triangle $P_{j-1}P_{j}P_{j+1}$ such that $[P_{j-1}P_{j}P_{j+1}]\ge\alpha$ . Analogously we can prove that there is a triangle such that $[P_{j-1}P_{j}P_{j+1}]\le\alpha$ .

(This proof is not mine, its from a teacher)

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a