A convex quadrillateral $ABCD$ is given and the intersection point of the diagonals is denoted by $O$. Given that the perimeters of the triangles $ABO, BCO, CDO,ADO$ are equal, prove that $ABCD$ is a rhombus.
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Tags: geometry, rhombus, perimeter
logrange
31.12.2020 11:53
WLOG BC≥AB≥CD≥AD BC≥CD implies BO≤DO (As, BC+BO+CO = CD+CO+DO) ...equation 1 Similarly, BC≥AB implies CO≤AO ...equation 2 and, AB≥CD implies OB+AO ≤ DO+CO...equation 3 By adding equation 1 and 2 we get BO+CO ≤AO+DO ...equation 4 Comparing equation 3 with equation 4 we get BO≤DO...equation 5 AD+OD+AO = AO+OB+AB(Given)...equation 6 AD≥AB (Comparing equation 6 with equation 5)...equation 7 AD=AB (By WLOG statement and equation 7)...equation 8. Putting this in equation 6 we get, OB=OD...equation 9 Putting OB=OD in equation 3 we get AO≤CO. Comparing this with equation 2 we get AO=CO...equation 10 This is a parallelogram (equation 9 and 10) in which adjacent sides are equal (equation 8). Thus, this is a rhombus. Hence, proved Hope this helps
greenturtle3141
31.12.2020 16:49
This was also on Russian Coffin problems. Here is a beautiful solution:
Note that $BC+CO = AB+AO$, so $BO$ is the perimeter bisector of $\triangle ABC$. This motivates considering of the excenter and the incenter. We similarly have $DO$ as the perimeter bisector of $\triangle ACD$.
Now let $P$ be the intouch-point of the incircle of $\triangle ABC$ with $AC$. $P$ must also be the intouch-point of the incircle of $\triangle ACD$. Let $X$ be the antipode of $P$ wrt the incircle of $\triangle ABD$, and let $Y$ be the antipode of $P$ wrt the incircle of $\triangle ACD$. Then clearly we have $X$, $Y$, $P$ collinear.
By a well-known lemma, we have $B$, $X$, $O$ collinear, and $D$, $Y$, $O$ collinear. But $B$, $O$, $D$ collinear, so $B$, $X$, $O$, $Y$, and $D$ are all collinear. So $X$, $O$, $Y$ collinear. So $XY$ intersects $AC$ at $O$. But $XY$ intersects $AC$ at $P$. Thus $O=P$.
With $O=P$, we can conclude that $\triangle ABC$ is isosceles with $BC = BA$. Rotate the argument thrice and we can conclude.