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Hi!

While participating in an online course in competition mathematics, I ran into a problem that I had some difficulty with. After struggling with it for a long time, I still have not reached the solution. I have attempted using a combination of coordinate geometry and heron's formula, but that has not worked. Apparently, trigonometry should be applied to solve this problem. However, I have not been able to find a particularly useful way to apply trigonometry here, yet.

The problem is as follows:

Two circles are internally tangent at a point T and have radii of 1 and 3. The maximum possible area for a triangle with one vertex at T, another vertex on the small circle, and the third on the large circle can be expressed in the form a*sqrt(b)/c, where a,b, and c are positive integers,  b is not divisible by the square of any prime, and a and c are relatively prime. Find a+b+c.

Any help would be greatly appreciated!

Thank you in advance!

 Sep 7, 2019
edited by jeyat  Sep 7, 2019

Best Answer 

 #1
avatar+23579 
+2

Two circles are internally tangent at a point T and have radii of 1 and 3.
The maximum possible area for a triangle with one vertex at T,
another vertex on the small circle,
and the third on the large circle can be expressed in the form a*sqrt(b)/c,
where a,b, and c are positive integers,
b is not divisible by the square of any prime,
and a and c are relatively prime.
Find a+b+c.

 

\(\text{Let $P_1$ point on small circle } \\ \text{Let $P_2$ point on large circle } \\ \text{Let $A$ area of the triangle $T-P_1-P_2$ }\)

 

\(\begin{array}{|rcll|} \hline T &=& \begin{pmatrix} 0\\ 0 \end{pmatrix} \\ \hline P_1 = \begin{pmatrix} x_1\\ y_1 \end{pmatrix} &=& \begin{pmatrix} r_1\cos(\varphi_1)+r_1 \\ r_1\sin(\varphi_1) \end{pmatrix} \\\\ &=& \begin{pmatrix} 1*\cos(\varphi_1)+1 \\ 1*\sin(\varphi_1) \end{pmatrix} \\\\ \begin{pmatrix} x_1\\ y_1 \end{pmatrix} &=& \begin{pmatrix} \cos(\varphi_1)+1 \\ \sin(\varphi_1) \end{pmatrix} \\ \hline P_2 = \begin{pmatrix} x_2\\ y_2 \end{pmatrix} &=& \begin{pmatrix} r_2\cos(\varphi_2)+r_2 \\ r_2\sin(\varphi_2) \end{pmatrix} \\\\ &=& \begin{pmatrix} 3\cos(\varphi_2)+3 \\ 3\sin(\varphi_2) \end{pmatrix} \\\\ \begin{pmatrix} x_2\\ y_2 \end{pmatrix} &=& \begin{pmatrix} 3(\cos(\varphi_2)+1) \\ 3\sin(\varphi_2) \end{pmatrix} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline 2A &=& |P_1\times P_2| \\ 2A &=& x_1y_2-y_1x_2 \\ &=& (\cos(\varphi_1)+1)3\sin(\varphi_2)-\sin(\varphi_1)3(\cos(\varphi_2)+1) \\ 2A&=& 3\Big( (\cos(\varphi_1)+1)\sin(\varphi_2)-\sin(\varphi_1)(\cos(\varphi_2)+1) \Big) \\\\ \mathbf{A} = f(\varphi_1,\varphi_2) &=& \mathbf{\dfrac{3}{2}\Big( (\cos(\varphi_1)+1)\sin(\varphi_2)-\sin(\varphi_1)(\cos(\varphi_2)+1) \Big)} \\ \hline \end{array}\)

 

maximum:

\(\begin{array}{|rcll|} \hline f_{\varphi_1} &=& \dfrac{\partial f(\varphi_1,\varphi_2)}{\partial \varphi_1}=0 \\ &=& \mathbf{\dfrac{3}{2}\Big( -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)(1+\cos(\varphi_2)) \Big)} \quad & (1) \\ \hline f_{\varphi_2} &=& \dfrac{\partial f(\varphi_1,\varphi_2)}{\partial \varphi_2}=0 \\ &=& \mathbf{\dfrac{3}{2}\Big( \cos(\varphi_2)(1+\cos(\varphi_1))-\sin(\varphi_1)(-\sin(\varphi_2)) \Big)} \quad & (2) \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1) & \dfrac{3}{2}\Big( -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)(1+\cos(\varphi_2)) \Big) &=& 0 \quad | \quad : \dfrac{3}{2} \\ & -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)(1+\cos(\varphi_2)) &=& 0 \\ & -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)-\cos(\varphi_1)\cos(\varphi_2) &=& 0 \\\\ (2) & \dfrac{3}{2}\Big( \sin(\varphi_1)\sin(\varphi_2)+\cos(\varphi_2)(1+\cos(\varphi_1)) \Big) &=& 0 \quad | \quad : \dfrac{3}{2} \\ & \sin(\varphi_1)\sin(\varphi_2) +\cos(\varphi_2)(1+\cos(\varphi_1)) &=& 0 \\ & \sin(\varphi_1)\sin(\varphi_2) +\cos(\varphi_2)+\cos(\varphi_1)\cos(\varphi_2) &=& 0 \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline (1) & -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)-\cos(\varphi_1)\cos(\varphi_2) &=& 0 \\ (2) & \sin(\varphi_1)\sin(\varphi_2) +\cos(\varphi_2)+\cos(\varphi_1)\cos(\varphi_2) &=& 0 \\ \hline (1)+(2): & -\cos(\varphi_1)+\cos(\varphi_2) &=& 0 \\ & \cos(\varphi_2) &=& \cos(\varphi_1) \\ & \varphi_2 &=& \pm \arccos(\cos\varphi_1) \\\\ & \varphi_2 &=& \varphi_1 \Rightarrow \text{minimum} \\\\ & \varphi_2 &=& -\varphi_1 \Rightarrow \text{maximum} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline (2) & \sin(\varphi_1)\sin(\varphi_2) +\cos(\varphi_2)+\cos(\varphi_1)\cos(\varphi_2) &=& 0 \quad | \quad \varphi_2 = -\varphi_1 \\ & \sin(\varphi_1)\sin(-\varphi_1) +\cos(-\varphi_1)+\cos(\varphi_1)\cos(-\varphi_1) &=& 0 \\ & -\sin(\varphi_1)\sin(\varphi_1) +\cos(\varphi_1)+\cos(\varphi_1)\cos(\varphi_1) &=& 0 \\ & \cos^2(\varphi_1)-\sin^2(\varphi_1)+\cos(\varphi_1)&=& 0 \quad | \quad \sin^2(\varphi_1) = 1-\cos^2(\varphi_1)\\ & \cos^2(\varphi_1)-\left(1-\cos^2(\varphi_1)\right)+\cos(\varphi_1)&=& 0 \\ & \cos^2(\varphi_1)-1+\cos^2(\varphi_1)+\cos(\varphi_1)&=& 0 \\ & 2\cos^2(\varphi_1)+\cos(\varphi_1)-1 &=& 0 \\\\ & \cos(\varphi_1) &=& \dfrac{-1\pm \sqrt{1-4\cdot 2\cdot(-1)}}{2\cdot 2} \\ & \cos(\varphi_1) &=& \dfrac{-1\pm \sqrt{9}}{4} \\ & \cos(\varphi_1) &=& \dfrac{-1\pm 3}{4} \\\\ &\cos(\varphi_1) &=&\dfrac{-1+ 3}{4} &\cos(\varphi_1) &=&\dfrac{1}{2} \\ & \varphi_1 &=& \arccos(\dfrac{1}{2}) \\ & \mathbf{\varphi_1} &=& \mathbf{60^\circ} \Rightarrow \text{ maximum} \\\\ &\cos(\varphi_1) &=&\dfrac{-1- 3}{4} &\cos(\varphi_1) &=& -1 \\ & \varphi_1 &=& \arccos(-1 ) \\ & \varphi_1 &=& 180^\circ \Rightarrow \text{ minimum} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \varphi_2 &=& -\varphi_1 \quad | \quad \mathbf{\varphi_1 = 60^\circ } \\ \mathbf{\varphi_2} &=& \mathbf{- 60^\circ} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{A} &=& \mathbf{\dfrac{3}{2}\Big( (\cos(\varphi_1)+1)\sin(\varphi_2)-\sin(\varphi_1)(\cos(\varphi_2)+1) \Big)} \quad | \quad \varphi_1 = 60^\circ,\ \varphi_2 = -60^\circ \\ &=&\dfrac{3}{2}\Big( (\cos(60^\circ)+1)\sin(-60^\circ)-\sin(60^\circ)(\cos(-60^\circ)+1) \Big) \\ &=&\dfrac{3}{2}\Big( (\cos(60^\circ)+1)\left(-\sin(60^\circ)\right) -\sin(60^\circ)(\cos(60^\circ)+1) \Big)\quad | \quad \cos(60^\circ) = \dfrac{1}{2},\ \sin(60^\circ)=\dfrac{\sqrt{3}}{2} \\ &=&\dfrac{3}{2}\Big( (\dfrac{1}{2}+1)\left(-\dfrac{\sqrt{3}}{2}\right) -\dfrac{\sqrt{3}}{2}(\dfrac{1}{2}+1) \Big) \\\\ &=&\dfrac{3}{2}\Big( -(\dfrac{1}{2}+1) \dfrac{\sqrt{3}}{2} -\dfrac{\sqrt{3}}{2}(\dfrac{1}{2}+1) \Big) \\\\ &=&\dfrac{3}{2}\Big( -2(\dfrac{1}{2}+1) \dfrac{\sqrt{3}}{2} \Big) \\\\ &=&\dfrac{3}{2}\Big( - (\dfrac{1}{2}+1) \sqrt{3} \Big) \\\\ &=&-\dfrac{3}{2}\Big(\dfrac{3}{2} \sqrt{3} \Big) \\\\ &=&-\dfrac{9\sqrt{3}}{4} \\\\ \mathbf{|A|} &=& \mathbf{\dfrac{9\sqrt{3}}{4}} \\\\ && \boxed{a=9,\ b=3,\ c=4 } \\\\ a+b+c &=& 9+3+4 \\ \mathbf{a+b+c} &=& \mathbf{16} \\ \hline \end{array}\)

 

laugh

 Sep 7, 2019
edited by heureka  Sep 8, 2019
edited by heureka  Sep 8, 2019
 #1
avatar+23579 
+2
Best Answer

Two circles are internally tangent at a point T and have radii of 1 and 3.
The maximum possible area for a triangle with one vertex at T,
another vertex on the small circle,
and the third on the large circle can be expressed in the form a*sqrt(b)/c,
where a,b, and c are positive integers,
b is not divisible by the square of any prime,
and a and c are relatively prime.
Find a+b+c.

 

\(\text{Let $P_1$ point on small circle } \\ \text{Let $P_2$ point on large circle } \\ \text{Let $A$ area of the triangle $T-P_1-P_2$ }\)

 

\(\begin{array}{|rcll|} \hline T &=& \begin{pmatrix} 0\\ 0 \end{pmatrix} \\ \hline P_1 = \begin{pmatrix} x_1\\ y_1 \end{pmatrix} &=& \begin{pmatrix} r_1\cos(\varphi_1)+r_1 \\ r_1\sin(\varphi_1) \end{pmatrix} \\\\ &=& \begin{pmatrix} 1*\cos(\varphi_1)+1 \\ 1*\sin(\varphi_1) \end{pmatrix} \\\\ \begin{pmatrix} x_1\\ y_1 \end{pmatrix} &=& \begin{pmatrix} \cos(\varphi_1)+1 \\ \sin(\varphi_1) \end{pmatrix} \\ \hline P_2 = \begin{pmatrix} x_2\\ y_2 \end{pmatrix} &=& \begin{pmatrix} r_2\cos(\varphi_2)+r_2 \\ r_2\sin(\varphi_2) \end{pmatrix} \\\\ &=& \begin{pmatrix} 3\cos(\varphi_2)+3 \\ 3\sin(\varphi_2) \end{pmatrix} \\\\ \begin{pmatrix} x_2\\ y_2 \end{pmatrix} &=& \begin{pmatrix} 3(\cos(\varphi_2)+1) \\ 3\sin(\varphi_2) \end{pmatrix} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline 2A &=& |P_1\times P_2| \\ 2A &=& x_1y_2-y_1x_2 \\ &=& (\cos(\varphi_1)+1)3\sin(\varphi_2)-\sin(\varphi_1)3(\cos(\varphi_2)+1) \\ 2A&=& 3\Big( (\cos(\varphi_1)+1)\sin(\varphi_2)-\sin(\varphi_1)(\cos(\varphi_2)+1) \Big) \\\\ \mathbf{A} = f(\varphi_1,\varphi_2) &=& \mathbf{\dfrac{3}{2}\Big( (\cos(\varphi_1)+1)\sin(\varphi_2)-\sin(\varphi_1)(\cos(\varphi_2)+1) \Big)} \\ \hline \end{array}\)

 

maximum:

\(\begin{array}{|rcll|} \hline f_{\varphi_1} &=& \dfrac{\partial f(\varphi_1,\varphi_2)}{\partial \varphi_1}=0 \\ &=& \mathbf{\dfrac{3}{2}\Big( -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)(1+\cos(\varphi_2)) \Big)} \quad & (1) \\ \hline f_{\varphi_2} &=& \dfrac{\partial f(\varphi_1,\varphi_2)}{\partial \varphi_2}=0 \\ &=& \mathbf{\dfrac{3}{2}\Big( \cos(\varphi_2)(1+\cos(\varphi_1))-\sin(\varphi_1)(-\sin(\varphi_2)) \Big)} \quad & (2) \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1) & \dfrac{3}{2}\Big( -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)(1+\cos(\varphi_2)) \Big) &=& 0 \quad | \quad : \dfrac{3}{2} \\ & -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)(1+\cos(\varphi_2)) &=& 0 \\ & -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)-\cos(\varphi_1)\cos(\varphi_2) &=& 0 \\\\ (2) & \dfrac{3}{2}\Big( \sin(\varphi_1)\sin(\varphi_2)+\cos(\varphi_2)(1+\cos(\varphi_1)) \Big) &=& 0 \quad | \quad : \dfrac{3}{2} \\ & \sin(\varphi_1)\sin(\varphi_2) +\cos(\varphi_2)(1+\cos(\varphi_1)) &=& 0 \\ & \sin(\varphi_1)\sin(\varphi_2) +\cos(\varphi_2)+\cos(\varphi_1)\cos(\varphi_2) &=& 0 \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline (1) & -\sin(\varphi_1)\sin(\varphi_2) -\cos(\varphi_1)-\cos(\varphi_1)\cos(\varphi_2) &=& 0 \\ (2) & \sin(\varphi_1)\sin(\varphi_2) +\cos(\varphi_2)+\cos(\varphi_1)\cos(\varphi_2) &=& 0 \\ \hline (1)+(2): & -\cos(\varphi_1)+\cos(\varphi_2) &=& 0 \\ & \cos(\varphi_2) &=& \cos(\varphi_1) \\ & \varphi_2 &=& \pm \arccos(\cos\varphi_1) \\\\ & \varphi_2 &=& \varphi_1 \Rightarrow \text{minimum} \\\\ & \varphi_2 &=& -\varphi_1 \Rightarrow \text{maximum} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline (2) & \sin(\varphi_1)\sin(\varphi_2) +\cos(\varphi_2)+\cos(\varphi_1)\cos(\varphi_2) &=& 0 \quad | \quad \varphi_2 = -\varphi_1 \\ & \sin(\varphi_1)\sin(-\varphi_1) +\cos(-\varphi_1)+\cos(\varphi_1)\cos(-\varphi_1) &=& 0 \\ & -\sin(\varphi_1)\sin(\varphi_1) +\cos(\varphi_1)+\cos(\varphi_1)\cos(\varphi_1) &=& 0 \\ & \cos^2(\varphi_1)-\sin^2(\varphi_1)+\cos(\varphi_1)&=& 0 \quad | \quad \sin^2(\varphi_1) = 1-\cos^2(\varphi_1)\\ & \cos^2(\varphi_1)-\left(1-\cos^2(\varphi_1)\right)+\cos(\varphi_1)&=& 0 \\ & \cos^2(\varphi_1)-1+\cos^2(\varphi_1)+\cos(\varphi_1)&=& 0 \\ & 2\cos^2(\varphi_1)+\cos(\varphi_1)-1 &=& 0 \\\\ & \cos(\varphi_1) &=& \dfrac{-1\pm \sqrt{1-4\cdot 2\cdot(-1)}}{2\cdot 2} \\ & \cos(\varphi_1) &=& \dfrac{-1\pm \sqrt{9}}{4} \\ & \cos(\varphi_1) &=& \dfrac{-1\pm 3}{4} \\\\ &\cos(\varphi_1) &=&\dfrac{-1+ 3}{4} &\cos(\varphi_1) &=&\dfrac{1}{2} \\ & \varphi_1 &=& \arccos(\dfrac{1}{2}) \\ & \mathbf{\varphi_1} &=& \mathbf{60^\circ} \Rightarrow \text{ maximum} \\\\ &\cos(\varphi_1) &=&\dfrac{-1- 3}{4} &\cos(\varphi_1) &=& -1 \\ & \varphi_1 &=& \arccos(-1 ) \\ & \varphi_1 &=& 180^\circ \Rightarrow \text{ minimum} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \varphi_2 &=& -\varphi_1 \quad | \quad \mathbf{\varphi_1 = 60^\circ } \\ \mathbf{\varphi_2} &=& \mathbf{- 60^\circ} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{A} &=& \mathbf{\dfrac{3}{2}\Big( (\cos(\varphi_1)+1)\sin(\varphi_2)-\sin(\varphi_1)(\cos(\varphi_2)+1) \Big)} \quad | \quad \varphi_1 = 60^\circ,\ \varphi_2 = -60^\circ \\ &=&\dfrac{3}{2}\Big( (\cos(60^\circ)+1)\sin(-60^\circ)-\sin(60^\circ)(\cos(-60^\circ)+1) \Big) \\ &=&\dfrac{3}{2}\Big( (\cos(60^\circ)+1)\left(-\sin(60^\circ)\right) -\sin(60^\circ)(\cos(60^\circ)+1) \Big)\quad | \quad \cos(60^\circ) = \dfrac{1}{2},\ \sin(60^\circ)=\dfrac{\sqrt{3}}{2} \\ &=&\dfrac{3}{2}\Big( (\dfrac{1}{2}+1)\left(-\dfrac{\sqrt{3}}{2}\right) -\dfrac{\sqrt{3}}{2}(\dfrac{1}{2}+1) \Big) \\\\ &=&\dfrac{3}{2}\Big( -(\dfrac{1}{2}+1) \dfrac{\sqrt{3}}{2} -\dfrac{\sqrt{3}}{2}(\dfrac{1}{2}+1) \Big) \\\\ &=&\dfrac{3}{2}\Big( -2(\dfrac{1}{2}+1) \dfrac{\sqrt{3}}{2} \Big) \\\\ &=&\dfrac{3}{2}\Big( - (\dfrac{1}{2}+1) \sqrt{3} \Big) \\\\ &=&-\dfrac{3}{2}\Big(\dfrac{3}{2} \sqrt{3} \Big) \\\\ &=&-\dfrac{9\sqrt{3}}{4} \\\\ \mathbf{|A|} &=& \mathbf{\dfrac{9\sqrt{3}}{4}} \\\\ && \boxed{a=9,\ b=3,\ c=4 } \\\\ a+b+c &=& 9+3+4 \\ \mathbf{a+b+c} &=& \mathbf{16} \\ \hline \end{array}\)

 

laugh

heureka Sep 7, 2019
edited by heureka  Sep 8, 2019
edited by heureka  Sep 8, 2019
 #2
avatar
+2

Hi heureka,

 

Thank you so much for your helpful response! Your approach is very interesting (and your answer is correct :) ).

 

Best,

jeyat

Guest Sep 8, 2019
 #3
avatar+23579 
+2

Thank you for your reply.

 

laugh

heureka  Sep 8, 2019

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