2005 IMO Shortlist Problems/G1


(Greece) In a triangle $\displaystyle ABC$ satisfying $\displaystyle AB + BC = 3AC$ the incircle has centre $\displaystyle I$ and touches the sides $\displaystyle AB$ and $\displaystyle BC$ at $\displaystyle D$ and $\displaystyle E$, respectively. Let $\displaystyle K$ and $\displaystyle L$ be the symmetric points of $\displaystyle D$ and $\displaystyle E$ with respect to $\displaystyle I$. Prove that the quadrilateral $\displaystyle ACKL$ is cyclic.

This was also Problem 2 of the second round of the 2006 Poland Math Olympiad and Problem 6 of the final round of the 2006 Costa Rica Math Olympiad.

Remark. The converse of this problem is also true, i.e., if $\displaystyle ACKL$ is cyclic, then $\displaystyle AB + BC = 3CA$. This can be proven easily along the lines of the first solution.


Solution 1

Let $\displaystyle BI$ intersect the circumcircle of $\displaystyle ABC$ again at $\displaystyle P$; let $\displaystyle M$ be the midpoint of $\displaystyle AC$. We note that $\displaystyle PM$ is the perpendicular bisector of $\displaystyle AC$, as it passes through the midpoint of segment $\displaystyle AC$ and the midpoint of the arc $\displaystyle AC$. We note that the condition $\displaystyle a+c = 3b$ is equivalent to the condition $\displaystyle AD = AE = (s-b) = 2b$.

Since $\angle MCP = \angle ACP = \angle ABP = \angle DBI$, and $\angle CMP$ and $\angle BDI$ are both right angles, triangles $\displaystyle MCP$, $\displaystyle DBI$ are similar.

Let $\displaystyle N$ be the projection of $\displaystyle P$ onto the line $\displaystyle IK$. We note that $\angle PNI$ and $\angle BDI$ are both right angles, and ${} \angle PIN = \angle BID$, so triangles $\displaystyle NPI, DBI$ are similar. Thus triangles $\displaystyle MCP, NPI$ are similar. But we note that by measures of intercepted arcs, $\angle ICP = \frac{\angle A + \angle C}{2} = \angle PIC$, so $\displaystyle PI = PC$ (in fact, this is a known result) and triangles $\displaystyle MCP, NPI$ are congruent. This means that $NI = MP = \frac{1}{2} DI = \frac{1}{2} KI$, so $\displaystyle PN$ is the perpendicular bisector of $\displaystyle KI$, and $\displaystyle PK = PI = CP$. By symmetry, $\displaystyle PA= PC = PK = PL$, so $\displaystyle A , C, K, L$ all lie on a circle centered at $\displaystyle P$.

Solution 2

Let the bisector of angle $\displaystyle B$ meet $\displaystyle DE$ at $\displaystyle B'$, $\displaystyle KL$ at $\displaystyle B''$, $\displaystyle AC$ at $\displaystyle Q$, and the circumcircle of $\displaystyle ABC$ a second time at $\displaystyle P$. Let $\displaystyle T$ be the midpoint of $\displaystyle AC$, and let the incircle of $\displaystyle ABC$ touch $\displaystyle AC$ at $\displaystyle F$. We note that $\displaystyle PT$ is the perpendicular bisector of $\displaystyle AC$, since this passes both through the midpoint of arc $\displaystyle APC$ and through the midpoint of $\displaystyle AC$. The condition $\displaystyle a+c = 3b$ is equivalent to the condition $\displaystyle AD= AE = (s-b) = b$.

Now, we have $\angle TAP = \angle CAP = \angle CBP = \angle EBI$, and $\angle ATP = \pi/2 = \angle BEI$, so triangles $\displaystyle ATP, BEI$ are similar. Thus we have $PT = \frac{IE \cdot AT}{BE} = \frac{r\cdot \frac{1}{2}b}{b} = \frac{r}{2}$. Considering the power of the point $\displaystyle T$ to the circumcircle of $\displaystyle ABC$ with respect to lines $\displaystyle AC, PT$ gives us

$\frac{1}{4}b^2 = \frac{1}{2}r \left( 2R - \frac{1}{2}r \right)$,

where $\displaystyle R$ and $\displaystyle r$ are the circumradius and inradius of $\displaystyle ABC$.

Then, the Pythagorean theorem on triangle $\displaystyle BEI$ gives us

$\displaystyle BI^2 = BE^2 + IE^2 = b^2 + r^2 = 4Rr$.

If $\displaystyle H_b$ is the foot of the altitude from $\displaystyle B$ to $\displaystyle AC$, then we have $BH_b = \frac{2[ABC]}{b} = \frac{2rs}{b} = \frac{2r\cdot 2b}{b} = 4r$. Furthermore, triangles $\displaystyle BH_bQ, IFQ$ are homothetic about $\displaystyle Q$, so

$\frac{BI}{IQ} + 1 = \frac{BQ}{IQ} = \frac{BH_b}{IF} = \frac{4r}{r} = 4$.

It follows that $\displaystyle IQ = \frac{2}{3}\sqrt{Rr}$. The extended law of sines gives us $4Rr = \frac{abc}{[ABC]} \cdot r = \frac{abc}{rs} \cdot r = \frac{abc}{2b} = \frac{ac}{2}$. By the angle bisector theorem, $AQ = \frac{c}{3}; CQ = \frac{a}{3}$. Considering the power of the point $\displaystyle Q$ with respect to the circumcircle of $\displaystyle ABC$ and lines $\displaystyle AC, BP$ gives us

$PT = \frac{AT \cdot CT}{AT} = \frac{\frac{ac}{9}}{(2+ 2/3)\sqrt{Rr}} = \sqrt{Rr}/3$.

Now, since $\displaystyle KB''I$ is the reflection of $\displaystyle DB'I$ about $\displaystyle I$, the two triangles are congruent. Since $\displaystyle DB'I, BDI$ are right triangles with a common acute angle, they are similar. Specifically,

$KB'' = DB' = \frac{BD \cdot DI}{BI} = \frac{b\cdot r}{\sqrt{Rr}}$,


$IB'' = IB' = \frac{ID \cdot DI}{BI} = \frac{r^2}{\sqrt{Rr}}$.


$PB'' = PI - B''I = \sqrt{Rr} - \frac{r^2}{\sqrt{Rr}}$.

Finally, applying the Pythagorean theorem to triangle $\displaystyle KB''P$, we have

$PK^2 = PB''^2 + KB''^2 = \left(\sqrt{Rr} - \frac{r^2}{\sqrt{Rr}}\right)^2 + \frac{b^2r}{R} = Rr -r^2 + \frac{r^3}{R} + \frac{r(4R-r)\cdot r}{R} = Rr$.

By symmetry, $PL = PK = \sqrt{Rr} = PA = PC$, so $\displaystyle A, C, K, L$ all lie on the circle with center $\displaystyle P$ and radius $\sqrt{Rr}$.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.