Difference between revisions of "Concurrence/Problems"

Revision as of 22:55, 28 November 2013

Introductory

Are the lines $y=2x+2$ , $y=3x+1$ , and $y=5x-1$ concurrent? If so, find the the point of concurrency.

Solution

If the points are concurrent, then they meet at one and only one point. We find where two of them meet:

$2x+2=3x+1 \Rightarrow x=1 \Rightarrow y=4$

We plug those into the third equation:

$5*1-1=4$

Therefore, $y=5x-1$ goes through the intersection of $y=2x+2$ and $y=3x+1$ , and those three lines are concurrent at $(1,4)$ .

Intermediate

See 1992 AIME Problems/Problem 14

Olympiad

Hallie is teaching geometry to Warren. She tells him that the three medians, the three angle bisectors, and the three altitudes of a triangle each meet at a point (the centroid, incenter, and orthocenter respectively). Warren gets a little confused and draws a certain triangle ABC along with the median from vertex A, the altitude from vertex B, and the angle bisector from vertex C. Hallie is surprised to see that the three segments meet at a point anyway! She notices that all three sides measure an integer number of inches, that the side lengths are all distinct, and that the side across from vertex C is 13 inches in length. How long are the other two sides?

Solution

(Letting $D$ be opposite of $A$ , and so forth) By Ceva's Theorem, we have

$\frac{AF}{BF} \cdot \frac{BD}{CD} \cdot \frac{CE}{AE} = 1$

The given tells us that $BD=CD$ , and by the Angle Bisector Theorem $\frac{AF}{BF} = \frac{AC}{BC}$ , so:

$\frac{AC}{BC} \cdot \frac{CE}{AE} = 1$

From here, we apply a bit of number theory to find that the answer is a $13-12-15$ triangle:

First, we derive a formula for the length of each line segment that an altitude from a vertex $B$ to a side $b$ divides $b$ into. WLOG, assuming that the line segment is $AE$ : $AB^2=BE^2+AE^2, BC^2=BE^2+CE^2$ Therefore: $AB^2-AE^2=BC^2-CE^2$ $AB^2-AE^2=BC^2-(AC-AE)^2$ $AB^2-AE^2=BC^2-AC^2-AE^2+2(AC)(AE)$ $AB^2=BC^2-AC^2+2AC(AE)$ $\frac{AB^2+AC^2-BC^2}{2(AC)}=AE$ $\frac{13^2+AC^2-BC^2}{2*AC}=AE$ To find the similar formula for $CE$ , we just switch the signs of $BC^2$ and $13^2$ . Returning to our original equation:

$\frac{AC}{BC} \cdot \frac {-13^2+AC^2+BC^2}{13^2+AC^2-BC^2} = 1$ $-AC*13^2+AC^3+AC*BC^2 = 13^2BC+AC^2*BC-BC^3$ $AC^3 - AC^2*BC +AC*BC^2 + BC^3= 13^2(BC+AC)$

Template:Incomplete

@@ Line 25: / Line 25: @@
 <cmath>\frac{AC}{BC} \cdot \frac{CE}{AE} = 1</cmath>
-From here, we apply a bit of [[number theory]] to find that the answer is a <math>13-12-15</math> triangle.
+From here, we apply a bit of [[number theory]] to find that the answer is a <math>13-12-15</math> triangle:
+First, we derive a formula for the length of each line segment that an altitude from a vertex <math>B</math> to a side <math>b</math> divides <math>b</math> into.
+WLOG, assuming that the line segment is <math>AE</math>:
+<cmath> AB^2=BE^2+AE^2, BC^2=BE^2+CE^2</cmath>
+Therefore:
+<cmath> AB^2-AE^2=BC^2-CE^2</cmath>
+<cmath> AB^2-AE^2=BC^2-(AC-AE)^2</cmath>
+<cmath> AB^2-AE^2=BC^2-AC^2-AE^2+2(AC)(AE)</cmath>
+<cmath> AB^2=BC^2-AC^2+2AC(AE)</cmath>
+<cmath> \frac{AB^2+AC^2-BC^2}{2(AC)}=AE</cmath>
+<cmath> \frac{13^2+AC^2-BC^2}{2*AC}=AE</cmath>
+To find the similar formula for <math>CE</math>, we just switch the signs of <math>BC^2</math> and <math>13^2</math>. Returning to our original equation:
+<cmath>\frac{AC}{BC} \cdot \frac {-13^2+AC^2+BC^2}{13^2+AC^2-BC^2} = 1</cmath>
+<cmath>-AC*13^2+AC^3+AC*BC^2 = 13^2BC+AC^2*BC-BC^3</cmath>
+<cmath>AC^3 - AC^2*BC +AC*BC^2 + BC^3= 13^2(BC+AC)</cmath>
 {{incomplete|solution}}

Page

Toolbox

Search

Difference between revisions of "Concurrence/Problems"

Revision as of 22:55, 28 November 2013

Contents

Introductory

Solution

Intermediate

Olympiad

Solution