During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

Difference between revisions of "Ceva's theorem/Problems"

(Created page with "==Introductory== ===I1=== ====Problem==== Suppose <math>AB, AC</math>, and <math>BC</math> have lengths <math>13, 14</math>, and <math>15</math>, respectively. If <math>\frac...")
 
 
Line 6: Line 6:
 
If <math>BD = x</math> and <math>DC = y</math>, then <math>10x = 40y</math>, and <math>{x + y = 15}</math>.  From this, we find <math>x = 12</math> and <math>y = 3</math>.
 
If <math>BD = x</math> and <math>DC = y</math>, then <math>10x = 40y</math>, and <math>{x + y = 15}</math>.  From this, we find <math>x = 12</math> and <math>y = 3</math>.
  
''[[Ceva's_Theorem | Back to main article]]''
+
''[[Ceva's_theorem | Back to main article]]''

Latest revision as of 15:16, 9 May 2021

Introductory

I1

Problem

Suppose $AB, AC$, and $BC$ have lengths $13, 14$, and $15$, respectively. If $\frac{AF}{FB} = \frac{2}{5}$ and $\frac{CE}{EA} = \frac{5}{8}$, find $BD$ and $DC$.

Solution

If $BD = x$ and $DC = y$, then $10x = 40y$, and ${x + y = 15}$. From this, we find $x = 12$ and $y = 3$.

Back to main article

Invalid username
Login to AoPS