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# Geometry

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Let $$\triangle MBT$$ be a triangle with MB = 4 and MT = 7. Furthermore, let circle ω be a circle with center O which is tangent to MB at B and MT at some point on segment MT. Given OM = 6 and ω intersect BT at I $$\neq$$ B, find the length of TI.

Jun 15, 2020

#1
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Let $$\triangle MBT$$ be a triangle with MB = 4 and MT = 7.
Furthermore, let circle $$\omega$$ be a circle with center O which is tangent to MB at B and MT at some point on segment MT.
Given OM = 6 and $$\omega$$ intersect BT at $$I \ne B$$, find the length of $$TI$$.

$$\text{Let TI = \color{red}x}$$

$$\begin{array}{|rcll|} \hline \mathbf{r=\ ?} \\ \hline 4^2+r^2 &=& 6^2 \\ r^2 &=& 6^2 -4^2 \\ r^2 &=& 36-16 \\ r^2 &=& 20 \\ r^2 &=& 4*5 \\ \mathbf{r} &=& \mathbf{2\sqrt{5}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \cos(A) &=& \dfrac{4}{6} \\\\ \cos(A) &=& \dfrac{2}{3} \\\\ \mathbf{\cos^2(A)} &=& \mathbf{\dfrac{4}{9}} \\\\ \cos(2A) &=& \cos^2(A)-\sin^2(a) \\ \cos(2A) &=& \cos^2(A)-\left(1- cos^2(A) \right) \\ \mathbf{\cos(2A)} &=& \mathbf{2\cos^2(A)-1} \\\\ \cos(2A) &=& 2*\dfrac{4}{9}-1 \\\\ \cos(2A) &=& \dfrac{8}{9}-1 \\\\ \cos(2A) &=& \dfrac{8}{9}-\dfrac{9}{9} \\\\ \mathbf{\cos(2A)} &=& \mathbf{-\dfrac{1}{9}} \\ \hline \end{array}$$

cos-rule:

$$\begin{array}{|rcll|} \hline BT^2 &=& 4^2+7^2-2*4*7*\cos(2A) \\ BT^2 &=& 16+49-56\cos(2A) \quad | \quad \mathbf{\cos(2A)=-\dfrac{1}{9}} \\ BT^2 &=& 16+49+56*\dfrac{1}{9} \\ BT^2 &=& 65+\dfrac{56}{9} \\ BT^2 &=& \dfrac{9*65+56}{9} \\ BT^2 &=& \dfrac{641}{9} \\ \mathbf{BT} &=& \mathbf{\dfrac{\sqrt{641}}{3}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{x=\ ?} \\ \hline 3^2 &=& x(x+BI) \quad | \quad BI = BT -x \\ 9 &=& x(x+BT -x) \\ 9 &=& x*BT \\ x &=& \dfrac{9}{BT} \quad | \quad \mathbf{BT=\dfrac{\sqrt{641}}{3}} \\\\ x &=& \dfrac{9}{\dfrac{\sqrt{641}}{3}} \\\\ x &=& \dfrac{3*9}{ \sqrt{641} } \\\\ x &=& \dfrac{27}{ \sqrt{641} } \\\\ x &=& \dfrac{27}{ 25.3179778023 } \\\\ \mathbf{x} &=& \mathbf{1.06643588247} \\ \hline \end{array}$$

The length of TI is  $$\approx \mathbf{1}$$

Jun 15, 2020
edited by heureka  Jun 15, 2020