Difference between revisions of "2021 AMC 12B Problems/Problem 17"
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==Solution== | ==Solution== | ||
− | Without loss 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 | 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 + 2\cdot\frac rs + 4\cdot\frac sr.</cmath> | <cmath>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.</cmath> | ||
− | Thus, the ratio <math>\rho := \tfrac rs</math> satisfies <math>\rho + 2\rho^{-1} = 4</math>; solving yields <math>\rho = \boxed{2+\sqrt 2\ | + | Thus, the ratio <math>\rho := \tfrac rs</math> satisfies <math>\rho + 2\rho^{-1} = 4</math>; solving yields <math>\rho = \boxed{\textbf{(B)}\: 2+\sqrt{2}}</math>. |
+ | |||
+ | (Observe that the given areas of <math>3</math> and <math>5</math> are irrelevant to the ratio <math>\frac{AB}{CD}</math>.) | ||
==Solution 2== | ==Solution 2== | ||
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Let <math>b_1</math> be the bottom base, <math>b_2</math> be the top base, <math>h_1</math> be the height of the bottom triangle, <math>h_2</math> be the height of the top triangle. | Let <math>b_1</math> be the bottom base, <math>b_2</math> be the top base, <math>h_1</math> be the height of the bottom triangle, <math>h_2</math> be the height of the top triangle. | ||
Thus, <math>b_1h_1 = 8, b_2h_2 = 4, (b_1+b_2)(h_1+h_2) = 28,</math> so <math>b_1h_2 + b_2h_1 = 16.</math> Let <math>b_2 = 1, h_2 = 4,</math> so we get <math>b_1h_1 = 8, 4b_1+h_1 = 16.</math> | Thus, <math>b_1h_1 = 8, b_2h_2 = 4, (b_1+b_2)(h_1+h_2) = 28,</math> so <math>b_1h_2 + b_2h_1 = 16.</math> Let <math>b_2 = 1, h_2 = 4,</math> so we get <math>b_1h_1 = 8, 4b_1+h_1 = 16.</math> | ||
− | This gives us a quadratic in <math>b_1,</math> ie. <math>4b_1^2+8=16b_1,</math> so <math>b_1 = \boxed{2+\sqrt{2}}.</math> | + | This gives us a quadratic in <math>b_1,</math> ie. <math>4b_1^2+8=16b_1,</math> so <math>b_1 = \boxed{\textbf{(B)}\: 2+\sqrt{2}}.</math> |
- Solution by MathAwesome123, added by ccx09 | - Solution by MathAwesome123, added by ccx09 | ||
+ | |||
+ | |||
+ | ==Video Solution by Challenge 25== | ||
+ | https://www.youtube.com/watch?v=0IB_1K8Dta8 | ||
+ | |||
+ | ==Video Solution (🚀 Just 3 min 🚀)== | ||
+ | https://youtu.be/XrIM3cdEk2k | ||
+ | |||
+ | <i>~Education, the Study of Everything </i> | ||
== Video Solution by OmegaLearn (Triangle Ratio and Trapezoid Area) == | == Video Solution by OmegaLearn (Triangle Ratio and Trapezoid Area) == |
Latest revision as of 11:58, 10 September 2023
Contents
[hide]Problem
Let be an isoceles trapezoid having parallel bases and with Line segments from a point inside to the vertices divide the trapezoid into four triangles whose areas are and starting with the triangle with base and moving clockwise as shown in the diagram below. What is the ratio
Solution
Without the loss of generality, let have vertices , , , and , with and . Also denote by the point in the interior of .
Let and be the feet of the perpendiculars from to and , respectively. Observe that and . Now using the formula for the area of a trapezoid yields Thus, the ratio satisfies ; solving yields .
(Observe that the given areas of and are irrelevant to the ratio .)
Solution 2
Let be the bottom base, be the top base, be the height of the bottom triangle, be the height of the top triangle. Thus, so Let so we get This gives us a quadratic in ie. so
- Solution by MathAwesome123, added by ccx09
Video Solution by Challenge 25
https://www.youtube.com/watch?v=0IB_1K8Dta8
Video Solution (🚀 Just 3 min 🚀)
~Education, the Study of Everything
Video Solution by OmegaLearn (Triangle Ratio and Trapezoid Area)
~ pi_is_3.14
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.