Question

Let $ABCD$ be an isosceles trapezoid having parallel bases $\overline{AB}$ and $\overline{CD}$ with $AB\gt CD.$ Line segments from a point inside $ABCD$ to the vertices divide the trapezoid into four triangles whose areas are $2, 3, 4,$ and $5$ starting with the triangle with base $\overline{CD}$ and moving clockwise as shown in the diagram below. What is the ratio $\dfrac{AB}{CD}?$

$3$

$2+\sqrt2$

$1+\sqrt6$

$2\sqrt3$

$3\sqrt2$

Solution:

Let $AB=a,$ $CD=b,$ and let the interior point be at heights $h_a$ from $AB$ and $h_b$ from $CD.$ The base triangles give $\tfrac12 a h_a=4$ and $\tfrac12 b h_b=2,$ so $a h_a=8$ and $b h_b=4.$

The total area is $2+3+4+5=14=\tfrac12(a+b)(h_a+h_b),$ so $(a+b)(h_a+h_b)=28.$ Expanding, $a h_a+b h_b+a h_b+b h_a=28,$ giving $a h_b+b h_a=16.$

Let $u=a h_b$ and $v=b h_a.$ Then $u+v=16$ and $uv=(a h_a)(b h_b)=32,$ so $u,v=8\pm 4\sqrt2.$

Finally $\dfrac{AB}{CD}=\dfrac{a}{b}=\dfrac{a h_b}{b h_b}=\dfrac{u}{4}=\dfrac{8+4\sqrt2}{4}=2+\sqrt2.$

Thus, the correct answer is B.

2021 AMC 12B Spring Problem 17

Solution:

Problem 17 in Other Years