$\angle{A}$ is the least angle in $\Delta{ABC}$. Point $D$ is on the arc $BC$ from the circumcircle of $\Delta{ABC}$. The perpendicular bisectors of the segments $AB,AC$ intersect the line $AD$ at $M,N$, respectively. Point $T$ is the meet point of $BM,CN$. Suppose that $R$ is the radius of the circumcircle of $\Delta{ABC}$. Prove that: \[ BT+CT\leq{2R}. \]
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Tags: geometry, circumcircle, trigonometry, inequalities, trapezoid, geometry proposed
Goutham
04.10.2010 11:27
Posted here.
kapilpavase
13.12.2014 09:45
Take $\angle{TBC}=x,\angle{TCB}=y$ We get easily that \[\angle{BTC}=2A\] By sine rule in$ \Delta ABC$, \[\dfrac{BC}{\sin{A}}=2R\] By sine rule in $\Delta BTC$, \[\dfrac{\sin{2A}}{BC}=\dfrac{\sin{x} +\sin{y}}{BT+CT}\] By jensen,\[\sin{x}+\sin{y}\le2\sin{\dfrac{x+y}{2}}=2\cos{A}\] So we get \[\dfrac{\sin{2A}}{BC}\le\dfrac{2\cos{A}}{BT+CT}\] \[\Rightarrow BT+CT\le\dfrac{BC}{\sin{A}}=2R\]
TelvCohl
13.12.2014 10:43
My solution: Let $ B'=BT \cap (ABC), C'=CT\cap (ABC) $ . Since $ \angle B'BC'=\angle MBA+\angle ACN=\angle BAC $ , so we get $ BC'B'C $ is an isosceles trapezoid and $ BT=TC' $ , hence $ BT+TC=C'T+TC=CC'=AD \leq 2R $ . Q.E.D
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Mahdi_Mashayekhi
22.12.2021 18:48
Let CT meet circumcircle at P. ∠BPT = ∠A and ∠PBT = ∠PBA + ∠ABM = ∠PCA + ∠BAM = ∠A ---> BT = PT so CT + BT = CP and CP is chord but 2R is diameter so 2R ≥ CT + BT.
rafaello
22.12.2021 21:30
By angle chase, $T$ lies on $(BOC)$. Moreover, $T$ lies on arc $BOC$ as $\angle ABC<90^\circ$ and $D$ lies on arc $BC$, not containing $A$. However, it is well-known that maximum of $TB+TC$ is achieved iff $T$ is the midpoint of arc (this can proved by calculus), hence equality occurs iff $T\equiv O$, which is equivalent to $AD$ passing through $O$.
nabodorbuco
13.01.2022 18:23
Hi there! just made a video about it. https://youtu.be/x9aSwgj0VBM
AgentC
06.02.2024 13:24
Hello, why should A be the least angle?