Difference between revisions of "Talk:2013 AMC 10B"
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Now you have AP = 4, CP = 3, PD = 2, and PE = 1.5 Apply A = 1/2bh to all of these triangles and add them up. You get 13.5 | Now you have AP = 4, CP = 3, PD = 2, and PE = 1.5 Apply A = 1/2bh to all of these triangles and add them up. You get 13.5 | ||
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| + | == Solution to problem 19 == | ||
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| + | The currently uploaded solution is correct until the part where it says <math> x=\sqrt{3}\/b/2 </math>. | ||
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| + | But since c=b+x but c is less than b, x has to be lower than zero, <math> x=-\sqrt{3}\/b/2 </math>. | ||
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| + | The root is <math> \frac{-b}{2a} </math>, equal to <math> \frac{-b}{2b+\sqrt{3}b} </math>. This is <math> \frac{-1}{2+\sqrt{3}} </math>, and multiplying by the conjugate gives the answer <math> -2+\sqrt{3} </math>. | ||
Latest revision as of 00:34, 26 February 2013
I think there is a simpler solution for number 16.
PE = 1.5, PD = 2, and DE = 2.5. This is a right triangle because of the Pythagorean Theorem (1.5^2 + 2^2 = 2.5^2). Since PED is a right triangle, ACP, CPD, and APE must be right triangles as well.
Since we know CE and AD are medians, they bust be bisected into 2:1 ratios. Therefore, CP = 3 (because PE * 2 = 3) and AP = 4 (because PD * 2 = 4).
Now you have AP = 4, CP = 3, PD = 2, and PE = 1.5 Apply A = 1/2bh to all of these triangles and add them up. You get 13.5
Solution to problem 19
The currently uploaded solution is correct until the part where it says 
.
But since c=b+x but c is less than b, x has to be lower than zero, 
.
The root is 
, equal to 
. This is 
, and multiplying by the conjugate gives the answer 
.
