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2022 SSMO Speed Round Problems/Problem 3: Difference between revisions

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Created page with "==Problem== Pigs like to eat carrots. Suppose a pig randomly chooses 6 letters from the set <math>\{c,a,r,o,t\}.</math> Then, the pig randomly arranges the 6 letters to form a..."
 
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==Problem==
==Problem==
Pigs like to eat carrots. Suppose a pig randomly chooses 6 letters from the set <math>\{c,a,r,o,t\}.</math> Then, the pig randomly arranges the 6 letters to form a "word". If the 6 letters don't spell carrot, the pig gets frustrated and tries to spell it again (by rechoosing the 6 letters and respelling them). What is the expected number of tries it takes for the pig to spell "carrot"?
 
Let <math>ABCD</math> be a parallelogram such that <math>E</math> is a point on <math>CD</math> such that <math>\frac{CE}{DE}=\frac{2}{3}.</math> Suppose that <math>BE</math> and <math>AC</math> intersect at <math>F.</math> If the area of triangle <math>AEF</math> is <math>15,</math> find the area of <math>ABCD</math>.


==Solution==
==Solution==
We first find the chance of the pig spelling "carrot" correctly in one try.
We have
 
<cmath>\begin{align*}
===Solution 1a===
[AEF] &= [AEC]-[CEF]\\
First, out of the <math>5^6</math> ways to choose the letters, only <math>\frac{6!}{2}</math> of them have the same letters as the word carrot. Then, given that the pig has chosen the words correctly, only <math>1</math> out of the <math>\frac{6!}{2}</math> ways to spell the word correctly.
&= \frac{CE}{CD}\cdot[ACD]-\frac{CE}{CD}\cdot\frac{CE}{CE+AB}\cdot[ACD]\\
 
&= \frac{CE}{CE+DE}\cdot[ACD]-\frac{CE}{CE+DE}\cdot\frac{CE}{CE+CD}\cdot[ACD]\\
The probability is thus
&= \frac{2}{2+3}\cdot[ACD]-\frac{2}{2+3}\cdot\frac{CE}{CE+(CE+DE)}\cdot[ACD]\\
<cmath>
&= \frac{2}{5}\cdot[ACD]-\frac{2}{5}\cdot\frac{2}{2+(2+3)}\cdot[ACD]\\
    \frac{\frac{6!}{2}}{5^6} \cdot \frac{1}{\frac{6!}{2}} = \frac{1}{5^6}
&= \left(\frac{2}{5}-\frac{2}{5}\cdot\frac{2}{7}\right)[ACD]\\
</cmath>
&= \frac{2}{7}[ACD] = \frac{2}{7}\cdot\frac{1}{2}\cdot[ABCD] = \frac{1}{7}[ABCD]\implies\\
 
[ABCD] &= 7[AEF] = 7\cdot15 = \boxed{105}.
===Solution 1b===
\end{align*}</cmath>
 
Considering each letter position individually, it is equally likely to be any
of the <math>5</math> possible letters. Thus, for each letter in carrot there is a
<math>\frac{1}{5}</math> chance the pig spells the letter in that position correctly.
The answer is thus <math>\frac{1}{5^6}</math>.
 
Now let <math>x</math> be the expected number of turns required for the pig to
guess correctly.


We have that
~pinkpig
<cmath>
    x = 1 + \frac{5^6 - 1}{5^6} \cdot x
</cmath>
which implies that <math>x = \boxed{15625}</math>

Latest revision as of 16:18, 13 September 2025

Problem

Let $ABCD$ be a parallelogram such that $E$ is a point on $CD$ such that $\frac{CE}{DE}=\frac{2}{3}.$ Suppose that $BE$ and $AC$ intersect at $F.$ If the area of triangle $AEF$ is $15,$ find the area of $ABCD$.

Solution

We have \begin{align*} [AEF] &= [AEC]-[CEF]\\ &= \frac{CE}{CD}\cdot[ACD]-\frac{CE}{CD}\cdot\frac{CE}{CE+AB}\cdot[ACD]\\ &= \frac{CE}{CE+DE}\cdot[ACD]-\frac{CE}{CE+DE}\cdot\frac{CE}{CE+CD}\cdot[ACD]\\ &= \frac{2}{2+3}\cdot[ACD]-\frac{2}{2+3}\cdot\frac{CE}{CE+(CE+DE)}\cdot[ACD]\\ &= \frac{2}{5}\cdot[ACD]-\frac{2}{5}\cdot\frac{2}{2+(2+3)}\cdot[ACD]\\ &= \left(\frac{2}{5}-\frac{2}{5}\cdot\frac{2}{7}\right)[ACD]\\ &= \frac{2}{7}[ACD] = \frac{2}{7}\cdot\frac{1}{2}\cdot[ABCD] = \frac{1}{7}[ABCD]\implies\\ [ABCD] &= 7[AEF] = 7\cdot15 = \boxed{105}. \end{align*}

~pinkpig

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