AMC 10 Weekly Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 22 Medium

On side AB of triangle ABC with area S, a point P is chosen at random. What is the probability that the area of \triangle PBC is greater than \frac{S}{3}?

  • A.

    \frac{1}{6}

  • B.

    \frac{2}{3}

  • C.

    \frac{1}{4}

  • D.

    \frac{1}{3}

  • E.

    \frac{5}{6}

Answer:B

Let the event be A=\left\{ \text{the area of } \triangle PBC \text{ is greater than } \frac{S}{3} \right\}.

 

The sample space is the length of segment AB (see figure).

Since {S}_{\triangle PBC} > \frac{S}{3},

 

we have \frac{1}{2} \cdot BC \cdot PE > \frac{1}{3} \times \frac{1}{2} \cdot BC \cdot AD,

 

which simplifies to \frac{PE}{AD} > \frac{1}{3}.

 

Since PE \parallel AD, by the similarity \triangle BPE \sim \triangle BAD, we obtain \frac{BP}{AB} = \frac{PE}{AD} > \frac{1}{3}.

 

Thus, the geometric measure of event A corresponds to the length on segment AB.

Since AP = \frac{2}{3}AB, we get P(A) = \frac{AP}{AB} = \frac{2}{3}.

 

Therefore, the answer is \text{B}.

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