Difference between revisions of "2021 AMC 12B Problems/Problem 11"
Lopkiloinm (talk | contribs) (Created page with "==Problem 11== Triangle <math>ABC</math> has <math>AB=13,BC=14</math> and <math>AC=15</math>. Let <math>P</math> be the point on <math>\overline{AC}</math> such that <math>PC=...") |
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===Solution 1=== | ===Solution 1=== | ||
| − | Using Stewart's Theorem of <math>man+dad=bmb+cnc</math> calculate the cevian to be <math>8\sqrt{2}</math>. It then follows that the answer must also have a factor of the <math>\sqrt{2}</math>. Having eliminated 3 answer choices, we then proceed to draw a rudimentary somewhat accurate diagram of this figure. Drawing that, we realize that <math>6\sqrt2</math> is too small making out answer <math>\boxed{\textbf{(D) }12\sqrt2}</math> | + | Using Stewart's Theorem of <math>man+dad=bmb+cnc</math> calculate the cevian to be <math>8\sqrt{2}</math>. It then follows that the answer must also have a factor of the <math>\sqrt{2}</math>. Having eliminated 3 answer choices, we then proceed to draw a rudimentary somewhat accurate diagram of this figure. Drawing that, we realize that <math>6\sqrt2</math> is too small making out answer <math>\boxed{\textbf{(D) }12\sqrt2}</math> ~Lopkiloinm |
Revision as of 18:24, 11 February 2021
Problem 11
Triangle 
has 
and 
. Let 
be the point on 
such that 
. There are exactly two points 
and 
on line 
such that quadrilaterals 
and 
are trapezoids. What is the distance 
Solutions
Solution 1
Using Stewart's Theorem of 
calculate the cevian to be 
. It then follows that the answer must also have a factor of the 
. Having eliminated 3 answer choices, we then proceed to draw a rudimentary somewhat accurate diagram of this figure. Drawing that, we realize that 
is too small making out answer 
~Lopkiloinm
