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

Revision as of 14:09, 4 August 2021

Problem

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 the loss of generality, 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 $14 = \frac12\cdot XY\cdot (AB+CD) = \frac12\left(\frac 8r + \frac 4s\right)(r+s) = 6 + 2\cdot\frac rs + 4\cdot\frac sr.$ Thus, the ratio $\rho := \tfrac rs$ satisfies $\rho + 2\rho^{-1} = 4$ ; solving yields $\rho = \boxed{2+\sqrt 2\textbf{ (B)}}$ .

Solution 2

Let $b_1$ be the bottom base, $b_2$ be the top base, $h_1$ be the height of the bottom triangle, $h_2$ be the height of the top triangle. Thus, $b_1h_1 = 8, b_2h_2 = 4, (b_1+b_2)(h_1+h_2) = 28,$ so $b_1h_2 + b_2h_1 = 16.$ Let $b_2 = 1, h_2 = 4,$ so we get $b_1h_1 = 8, 4b_1+h_1 = 16.$ This gives us a quadratic in $b_1,$ ie. $4b_1^2+8=16b_1,$ so $b_1 = \boxed{2+\sqrt{2}}.$

- Solution by MathAwesome123, added by ccx09

Video Solution by OmegaLearn (Triangle Ratio and Trapezoid Area)

https://youtu.be/MpMdRI9wC54

~ pi_is_3.14

@@ Line 26: / Line 26: @@
 ==Solution==
-Without loss of generality let <math>\mathcal T</math> have vertices <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>, with <math>AB = r</math> and <math>CD = s</math>. Also denote by <math>P</math> the point in the interior of <math>\mathcal T</math>.
+Without the loss of generality, let <math>\mathcal T</math> have vertices <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>, with <math>AB = r</math> and <math>CD = s</math>. Also denote by <math>P</math> the point in the interior of <math>\mathcal T</math>.
 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

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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