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**Tangent Circles Finding solutions with both**

Law of Cosines and Stewart’s Theorem Inspired by (though not actually stolen from) problem #10 on the 2011 High School Purple Comet Competition With thanks to Luke Shimanuki for sharing his solution

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3 inches

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**AB = AC = BC = AP = BP = CP = 3 2 1 r + 1 r + 2 3 - r**

Find the following segments either as whole numbers or in terms of r (the radius of circle P) AB = AC = BC = AP = BP = CP = 3 2 1 r + 1 r + 2 3 - r

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**AB = AC = BC = AP = BP = CP = 3 2 1 r + 1 r + 2 3 - r**

Let us now focus on two of the triangles – APC and APB I will redraw and enlarge the two triangles on the next slide

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r + 1 3 - r r + 2 x 2 1 We will begin by using Law of Cosines with triangle APC to solve for angle PAC (x degrees)

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**Law of Cosines a2 = b2 + c2 – 2bcCos A Triangle PAC 2 1 r + 1 r + 2**

x Law of Cosines a2 = b2 + c2 – 2bcCos A Triangle PAC

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**Now we will use law of cosines a second time**

2 1 r + 1 r + 2 3 - r x Law of Cosines Triangle PAC Now we will use law of cosines a second time This time we will use it on Triangle PAB

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**Law of Cosines - Triangle PAB**

2 1 r + 1 r + 2 3 - r x Law of Cosines - Triangle PAB a2 = b2 + c2 – 2bcCos A

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3 inches

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**Allows you to calculate the length of a cevian.**

Stewart’s Theorem Allows you to calculate the length of a cevian. A cevian is a line segment that extends from a vertex of a polygon to it’s opposite side. Medians and altitudes are examples of “special” cevians in a triangle. The formula is:

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**We can rewrite the formula in a way that is easier to remember …**

Stewart’s Theorem A cevian is a line segment that extends from a vertex of a polygon to it’s opposite side. We can rewrite the formula in a way that is easier to remember … A man and his dad put a bomb in the sink ….

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**Let’s plug our values in and see what we get**

Stewart’s Theorem Let’s plug our values in and see what we get a = b = c = d = m = n = 3 r + 2 r + 1 3 – r 2 1

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a Stewart’s Theorem a = b = c = d = m = n = 3 r + 2 r + 1 3 – r 2 1

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