Difference between revisions of "2019 AMC 10B Problems/Problem 16"
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− | + | ==Problem== | |
+ | In <math>\triangle ABC</math> with a right angle at <math>C</math>, point <math>D</math> lies in the interior of <math>\overline{AB}</math> and point <math>E</math> lies in the interior of <math>\overline{BC}</math> so that <math>AC=CD,</math> <math>DE=EB,</math> and the ratio <math>AC:DE=4:3</math>. What is the ratio <math>AD:DB?</math> | ||
+ | |||
+ | <math>\textbf{(A) }2:3\qquad\textbf{(B) }2:\sqrt{5}\qquad\textbf{(C) }1:1\qquad\textbf{(D) }3:\sqrt{5}\qquad\textbf{(E) }3:2</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | Without loss of generality, let <math>AC = CD = 4</math> and <math>DE = EB = 3</math>. Let <math>\angle A = \alpha</math> and <math>\angle B = \beta = 90^{\circ} - \alpha</math>. As <math>\triangle ACD</math> and <math>\triangle DEB</math> are isosceles, <math>\angle ADC = \alpha</math> and <math>\angle BDE = \beta</math>. Then <math>\angle CDE = 180^{\circ} - \alpha - \beta = 90^{\circ}</math>, so <math>\triangle CDE</math> is a <math>3-4-5</math> triangle with <math>CE = 5</math>. | ||
+ | |||
+ | Then <math>CB = 5+3 = 8</math>, and <math>\triangle ABC</math> is a <math>1-2-\sqrt{5}</math> triangle. | ||
+ | |||
+ | In isosceles triangles <math>\triangle ACD</math> and <math>\triangle DEB</math>, drop altitudes from <math>C</math> and <math>E</math> onto <math>AB</math>; denote the feet of these altitudes by <math>P_C</math> and <math>P_E</math> respectively. Then <math>\triangle ACP_C \sim \triangle ABC</math> by AAA similarity, so we get that <math>AP_C = P_CD = \frac{4}{\sqrt{5}}</math>, and <math>AD = 2 \times \frac{4}{\sqrt{5}}</math>. Similarly we get <math>BD = 2 \times \frac{6}{\sqrt{5}}</math>, and <math>AD:DB = \boxed{\textbf{(A) } 2:3}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>AC=CD=4x</math>, and <math>DE=EB=3x</math>. (For this solution, <math>A</math> is above <math>C</math>, and <math>B</math> is to the right of <math>C</math>). Also let <math>\angle A = t^{\circ}</math>, so <math>\angle ACD = \left(180-2t\right)^{\circ}</math>, which implies <math>\angle DCB = \left(2t - 90\right)^{\circ}</math>. Similarly, <math>\angle B = \left(90-t\right)^{\circ}</math>, which implies <math>\angle BED = 2t^{\circ}</math>. This further implies that <math>\angle DEC = \left(180 - 2t\right)^{\circ}</math>. | ||
+ | |||
+ | Now we see that <math>\angle CDE = 180^{\circ} - \angle ECD - \angle DEC = 180^{\circ} - 2t^{\circ} + 90^{\circ} - 180^{\circ} + 2t^{\circ} = 90^{\circ}</math>. Thus <math>\triangle CDE</math> is a right triangle, with side lengths of <math>3x</math>, <math>4x</math>, and <math>5x</math> (by the Pythagorean Theorem, or simply the Pythagorean triple <math>3-4-5</math>). Therefore <math>AC=4x</math> (by definition), <math>BC=5x+3x = 8x</math>, and <math>AB=4\sqrt{5}x</math>. Hence <math>\cos{\left(2t^{\circ}\right)} = 2 \cos^{2}{t^{\circ}} - 1</math> (by the double angle formula), giving <math>2\left(\frac{1}{\sqrt{5}}\right)^2 - 1 = -\frac{3}{5}</math>. | ||
+ | |||
+ | By the Law of Cosines in <math>\triangle BED</math>, if <math>BD = d</math>, we have <cmath>\begin{split}&d^2 = (3x)^2+(3x)^2-2\cdot\frac{-3}{5}(3x)(3x) \\ \Rightarrow \ &d^2 = 18x^2 + \frac{54x^2}{5} = \frac{144x^2}{5} \\ \Rightarrow \ &d = \frac{12x}{\sqrt{5}}\end{split}</cmath> Now <math>AD = AB - BD = 4x\sqrt{5} - \frac{12x}{\sqrt{5}} = \frac{8x}{\sqrt{5}}</math>. Thus the answer is <math>\frac{\left(\frac{8x}{\sqrt{5}}\right)}{\left(\frac{12x}{\sqrt{5}}\right)} = \frac{8}{12} = \boxed{\textbf{(A) }2:3}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | WLOG, let <math>AC=CD=4</math>, and <math>DE=EB=3</math>. <math>\angle CDE = 180^{\circ} - \angle ADC - \angle BDE = 180^{\circ} - \angle DAC - \angle DBE = 90^{\circ}</math>. Because of this, <math>\triangle DEC</math> is a 3-4-5 right triangle. Draw the altitude <math>DF</math> of <math>\triangle DEC</math>. <math>DF</math> is <math>\frac{12}{5}</math> by the base-height triangle area formula. <math>\triangle ABC</math> is similar to <math>\triangle DBF</math> (AA). So <math>\frac{DF}{AC} = \frac{BD}{AB} = \frac35</math>. <math>DB</math> is <math>\frac35</math> of <math>AB</math>. Therefore, <math>AD:DB</math> is <math>\boxed{\textbf{(A) } 2:3}</math>. | ||
+ | |||
+ | ~Thegreatboy90 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/_0YaCyxiMBo | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Latest revision as of 01:18, 26 September 2020
Problem
In with a right angle at , point lies in the interior of and point lies in the interior of so that and the ratio . What is the ratio
Solution 1
Without loss of generality, let and . Let and . As and are isosceles, and . Then , so is a triangle with .
Then , and is a triangle.
In isosceles triangles and , drop altitudes from and onto ; denote the feet of these altitudes by and respectively. Then by AAA similarity, so we get that , and . Similarly we get , and .
Solution 2
Let , and . (For this solution, is above , and is to the right of ). Also let , so , which implies . Similarly, , which implies . This further implies that .
Now we see that . Thus is a right triangle, with side lengths of , , and (by the Pythagorean Theorem, or simply the Pythagorean triple ). Therefore (by definition), , and . Hence (by the double angle formula), giving .
By the Law of Cosines in , if , we have Now . Thus the answer is .
Solution 3
WLOG, let , and . . Because of this, is a 3-4-5 right triangle. Draw the altitude of . is by the base-height triangle area formula. is similar to (AA). So . is of . Therefore, is .
~Thegreatboy90
Video Solution
~IceMatrix
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 10 Problems and Solutions |
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