A quadrilateral $ABCD$ without parallel sidelines is circumscribed around a circle centered at $I$. Let $K, L, M$ and $N$ be the midpoints of $AB, BC, CD$ and $DA$ respectively. It is known that $AB \cdot CD = 4IK \cdot IM$. Prove that $BC \cdot AD = 4IL \cdot IN$.
Problem
Source: Sharygin CR 2019 P18 (Grade 10 - 11)
Tags: geometry
06.03.2019 15:58
lol I had disproved this, then they changed the problem statement
Attachments:
Problem 18.pdf (128kb)
06.03.2019 19:54
Just a counter-example for the problem before it was corrected: $AB=4,~BC=5,~CD=6,~DA=5.$ @below. Yes I too obtained it, but when I was typing my solution, I used a counter-example.
06.03.2019 19:58
Actually $AD=BC$ and $CD\neq AB$ is a contradiction.
06.03.2019 21:38
Did anyone try complex bash for it with the incircle as the unit circle ? I tried it , but I was stuck in between , so I just sent whatever I got
06.03.2019 21:40
SHREYAS333 wrote: Did anyone try complex bash for it with the incircle as the unit circle ? I tried it , but I was stuck in between , so I just sent whatever I got I had tried it but it was seeming pretty ugly after one point.
01.07.2020 14:02
What a nice problem! (The post should also state that no two sides of $ABCD$ are equal.) We're given that $\tfrac{2IK}{AB} \cdot \tfrac{2IM}{CD} = 1$. Recall that $\angle AIB + \angle CID = 180^\circ$. If $\angle AIB = \angle CID = 90^\circ$ then $\overline{AD} \parallel \overline{BC}$, a contradiction. Now let $I_1$ be the inverse of $I$ in $(CD)$. Then $\tfrac{2IK}{AB} = \tfrac{2I_1M}{CD}$ and $\angle AIB = \angle CI_1D$. Since $\angle AIB$ and $\angle CI_1D$ are not $90^\circ$, this is enough to establish $\triangle AIB \sim \triangle CI_1D$ or $\triangle AIB \sim \triangle DI_1C$. This similarity implies that \[\frac{IC}{ID} = \frac{I_1C}{I_1D} \in \left\{\frac{IA}{IB}, \frac{IB}{IA}\right\}.\]Thus, we have either $IA \cdot IC = IB \cdot ID$ or $IA \cdot ID = IB \cdot IC$. However, the latter implies $AD = BC$, which is not allowed. To see this, let $r$ denote the inradius of $ABCD$; then we have \[ \frac{IA \cdot IB}{AB} = \frac{2[IAB]}{AB \sin \angle AIB} = \frac{2r}{\sin \angle AIB} = \frac{2r}{\sin \angle CID} = \frac{2[ICD]}{CD \sin \angle CID} = \frac{IC \cdot ID}{CD}.\] It follows that $\boxed{IA \cdot IC = IB \cdot ID}$. Now, the finish is essentially the reverse of the above. Note that $\tfrac{IA}{ID} = \tfrac{IB}{IC}$ but $\angle AIB + \angle CID = 180^\circ$. Letting $I_2$ be the inverse of $I$ in $(BC)$, we have $\triangle AID \sim \triangle BI_2C$. Thus \[\angle IAN = \angle IAD = \angle I_2BC = \angle I_2BC = \angle I_2BL = \angle BIL\]and likewise $\angle AIN = \angle IBL$. It follows that $\triangle IAN \sim \triangle BIL$, and finally \[\frac{IN}{AN} = \frac{BL}{IL} \implies AN \cdot BL = IL \cdot IN \implies BC \cdot AD = 4IL \cdot IN\]as desired.