Presentation is loading. Please wait.

# Tangent Circles Finding solutions with both

## Presentation on theme: "Tangent Circles Finding solutions with both"— Presentation transcript:

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

3 inches

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

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

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)

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

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

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

3 inches

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:

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 ….

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

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

Download ppt "Tangent Circles Finding solutions with both"

Similar presentations

Ads by Google