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The problem:

A trapezoid $ABCD$ lies on the xy-plane. The slopes of lines $BC$ and $AD$ are both $\frac{1}{3}$, and the slope of line $AB$ is $-\frac{2}{3}$. Given that $AB = CD$ and $BC < AD$, the absolute value of the slope of line $CD$ can be expressed as $\frac{m}{n}$, where m and n are two relatively prime positive integers. Find $\boxed{m + n}.$

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I drew the diagram and then I labeled $B$ as the origin which was the bottom left point in my diagram. I then set a random value equal to $AB$ and let point $A$ be $(-3, 2)$.

I then drew the heights from $B$ to $AD$ and $C$ to $AD$ and called them $E$ and $F$ respectively. Since I let point A be $(-3, 2)$, $AE = 3$. Similarly, $DF = 3$. $EF = BC = a$ for some value $a$. I then realized that since $AD$ had slope $\frac{1}{3}$, I used some variables to find out the point $D$. But, it was too messy and I'm stuck on what to do next after finding $D$. I also realized that $BC$ had slope $\frac{1}{3}$ and that I could get some equation between $BC$ and $AD$ but I couldn't get it.

Can someone tell me if what I was doing is correct (variable bashing) or should I come up with a more elegant solution? And can someone also tell me what to do next after finding $D$?

Qi Zhu
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I am a person
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1 Answers1

1

You should use the formula $\displaystyle \small \tan \theta = \big|\frac{m_2-m_1}{1+m_1m_2}\big|$ to find angle between $\small AD$ and $\small AB$. We know $\small \angle BAD$ is acute as $\small BC \lt AD, BC \parallel AD$ and $\small AB = CD$. If $\small \angle BAD = \theta, $

$\displaystyle \small \tan \theta = \frac{1/3 - (-2/3)}{1 + (1/3) \cdot (-2/3)} = \frac{9}{7}$

Now as $\displaystyle \small \angle ADC = \angle BAD, \ \frac{9}{7} = \big |\frac{m_3 - 1/3}{1 + m_3 / 3} \big| \ $ where $m_3$ is slope of $CD$.

You get two values of $m_3$. One of them gives you trapezoid and the other gives you a parallelogram.

Can you take it from here?

Math Lover
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