Difference between revisions of "2019 AMC 10B Problems/Problem 16"
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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>. | 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} - | + | 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> | By the Law of Cosines in <math>\triangle BED</math>, if <math>BD = d</math>, we have <cmath> | ||
==Solution 3== | ==Solution 3== | ||
− | Draw | + | 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== | ==See Also== | ||
{{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}} | {{AMC10 box|year=2019|ab=B|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:18, 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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.