2021 AMC 12B Problems/Problem 17

Problem 17

Let $ABCD$ be an isoceles trapezoid having parallel bases $\overline{AB}$ and $\overline{CD}$ with $AB>CD.$ 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 $\frac{AB}{CD}?$ $[asy] unitsize(100); pair A=(-1, 0), B=(1, 0), C=(0.3, 0.9), D=(-0.3, 0.9), P=(0.2, 0.5), E=(0.1, 0.75), F=(0.4, 0.5), G=(0.15, 0.2), H=(-0.3, 0.5); draw(A--B--C--D--cycle, black); draw(A--P, black); draw(B--P, black); draw(C--P, black); draw(D--P, black); label("$A$",A,(-1,0)); label("$B$",B,(1,0)); label("$C$",C,(1,-0)); label("$D$",D,(-1,0)); label("$2$",E,(0,0)); label("$3$",F,(0,0)); label("$4$",G,(0,0)); label("$5$",H,(0,0)); dot(A^^B^^C^^D^^P); [/asy]$ $\textbf{(A)}\: 3\qquad\textbf{(B)}\: 2+\sqrt{2}\qquad\textbf{(C)}\: 1+\sqrt{6}\qquad\textbf{(D)}\: 2\sqrt{3}\qquad\textbf{(E)}\: 3\sqrt{2}$

Solution

Without loss let $\mathcal T$ have vertices $A$ , $B$ , $C$ , and $D$ , with $AB = r$ and $CD = s$ . Also denote by $P$ the point in the interior of $\mathcal T$ .

Let $X$ and $Y$ be the feet of the perpendiculars from $P$ to $AB$ and $CD$ , respectively. Observe that $PX = \tfrac 8r$ and $PY = \tfrac 4s$ . Now using the formula for the area of a trapezoid yields \[ 14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 4\cdot\frac rs + 2\cdot\frac sr. \]Thus, the ratio $\rho := \tfrac rs$ satisfies $2\rho + \rho^{-1} = 4$ ; solving yields $\rho = \boxed{2+\sqrt 2\textbf{ (B)}}$ .

Page

Toolbox

Search

2021 AMC 12B Problems/Problem 17

Problem 17

Solution