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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.

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Presentation on theme: "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."— Presentation transcript:

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

4 Let us now focus on two of the triangles – APC and APB AB = AC = BC = AP = BP = CP = r + 1 r r I will redraw and enlarge the two triangles on the next slide

5 2 1 r + 1r r x We will begin by using Law of Cosines with triangle APC to solve for angle PAC (x degrees)

6 2 1 r + 1 r r x a 2 = b 2 + c 2 – 2bcCos A Law of Cosines Triangle PAC

7 2 1 r + 1 r 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

8 2 1 r + 1 r r x a 2 = b 2 + c 2 – 2bcCos A Law of Cosines - Triangle PAB

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10 a 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:

11 a 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 ….

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

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

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