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Difference between revisions of "2021 AMC 12B Problems/Problem 17"

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Let <math>X</math> and <math>Y</math> be the feet of the perpendiculars from <math>P</math> to <math>AB</math> and <math>CD</math>, respectively. Observe that <math>PX = \tfrac 8r</math> and <math>PY = \tfrac 4s</math>. Now using the formula for the area of a trapezoid yields
 
Let <math>X</math> and <math>Y</math> be the feet of the perpendiculars from <math>P</math> to <math>AB</math> and <math>CD</math>, respectively. Observe that <math>PX = \tfrac 8r</math> and <math>PY = \tfrac 4s</math>. Now using the formula for the area of a trapezoid yields
 
<cmath>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.</cmath>
 
<cmath>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.</cmath>
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Thus, the ratio <math>\rho := \tfrac rs</math> satisfies <math>2\rho + \rho^{-1} = 4</math>; solving yields <math>\rho = \boxed{2+\sqrt 2\textbf{ (B)}}</math>. - djmathman
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Thus, the ratio <math>\rho := \tfrac rs</math> satisfies <math>2\rho + \rho^{-1} = 4</math>; solving yields <math>\rho = \boxed{2+\sqrt 2\textbf{ (B)}}</math>.

Revision as of 14:56, 11 February 2021

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 $\rho = \boxed{2+\sqrt 2\textbf{ (B)}}$.

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