Question

Triangle $ABC$ has $AB=13, BC=14,$ and $AC=15.$ Let $P$ be the point on $\overline{AC}$ such that $PC=10.$ There are exactly two points $D$ and $E$ on line $BP$ such that quadrilaterals $ABCD$ and $ABCE$ are trapezoids. What is the distance $DE?$

$\dfrac{42}{5}$

$6\sqrt2$

$\dfrac{84}{5}$

$12\sqrt2$

$18$

Solution:

Place $A=(0,0)$ and $C=(15,0).$ Then $B=\left(\tfrac{33}{5},\tfrac{56}{5}\right),$ and since $PC=10,$ $P=(5,0).$ Line $BP$ has slope $7,$ so it is $y=7(x-5).$

For $ABCD$ to be a trapezoid with $D$ on line $BP,$ take $CD\parallel AB.$ The line through $C$ parallel to $AB$ meets line $BP$ at $(1.8,-22.4).$

For $ABCE$ with $E$ on line $BP,$ take $AE\parallel BC.$ The line through $A$ parallel to $BC$ meets line $BP$ at $(4.2,-5.6).$

The distance is $\sqrt{(4.2-1.8)^2+(-5.6+22.4)^2}=\sqrt{2.4^2+16.8^2}=\sqrt{288}=12\sqrt2.$

Thus, the correct answer is D.

2021 AMC 12B Spring Problem 11

Solution:

Problem 11 in Other Years