Difference between revisions of "2020 AIME I Problems/Problem 1"
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== Problem == | == Problem == | ||
− | In <math>\triangle ABC</math> with <math>AB=AC,</math> point <math>D</math> lies strictly between <math>A</math> and <math>C</math> on side <math>\overline{AC},</math> and point <math>E</math> lies strictly between <math>A</math> and <math>B</math> on side <math>\overline{AB}</math> such that <math>AE=ED=DB=BC.</math> The degree measure of <math>\angle ABC</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math> | + | In <math>\triangle ABC</math> with <math>AB=AC,</math> point <math>D</math> lies strictly between <math>A</math> and <math>C</math> on side <math>\overline{AC},</math> and point <math>E</math> lies strictly between <math>A</math> and <math>B</math> on side <math>\overline{AB}</math> such that <math>AE=ED=DB=BC.</math> The degree measure of <math>\angle ABC</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math> |
== Solution 1== | == Solution 1== | ||
Line 26: | Line 26: | ||
If we set <math>\angle{BAC}</math> to <math>x</math>, we can find all other angles through these two properties: | If we set <math>\angle{BAC}</math> to <math>x</math>, we can find all other angles through these two properties: | ||
1. Angles in a triangle sum to <math>180^{\circ}</math>. | 1. Angles in a triangle sum to <math>180^{\circ}</math>. | ||
− | 2. The base angles of an | + | 2. The base angles of an isosceles triangle are congruent. |
Now we angle chase. <math>\angle{ADE}=\angle{EAD}=x</math>, <math>\angle{AED} = 180-2x</math>, <math>\angle{BED}=\angle{EBD}=2x</math>, <math>\angle{EDB} = 180-4x</math>, <math>\angle{BDC} = \angle{BCD} = 3x</math>, <math>\angle{CBD} = 180-6x</math>. Since <math>AB = AC</math> as given by the problem, <math>\angle{ABC} = \angle{ACB}</math>, so <math>180-4x=3x</math>. Therefore, <math>x = 180/7^{\circ}</math>, and our desired angle is <cmath>180-4\left(\frac{180}{7}\right) = \frac{540}{7}</cmath> for an answer of <math>\boxed{547}</math>. | Now we angle chase. <math>\angle{ADE}=\angle{EAD}=x</math>, <math>\angle{AED} = 180-2x</math>, <math>\angle{BED}=\angle{EBD}=2x</math>, <math>\angle{EDB} = 180-4x</math>, <math>\angle{BDC} = \angle{BCD} = 3x</math>, <math>\angle{CBD} = 180-6x</math>. Since <math>AB = AC</math> as given by the problem, <math>\angle{ABC} = \angle{ACB}</math>, so <math>180-4x=3x</math>. Therefore, <math>x = 180/7^{\circ}</math>, and our desired angle is <cmath>180-4\left(\frac{180}{7}\right) = \frac{540}{7}</cmath> for an answer of <math>\boxed{547}</math>. | ||
+ | |||
+ | See here for a video solution: https://youtu.be/4e8Hk04Ax_E | ||
==Solution 2== | ==Solution 2== | ||
Line 36: | Line 38: | ||
This tells us <math>\angle{BCA}=\angle{ABC}=3x</math> and <math>3x+3x+x=180</math>. | This tells us <math>\angle{BCA}=\angle{ABC}=3x</math> and <math>3x+3x+x=180</math>. | ||
Thus <math>x=\frac{180}{7}</math> and we want <math>\angle{ABC}=3x=\frac{540}{7}</math> to get an answer of <math>\boxed{547}</math>. | Thus <math>x=\frac{180}{7}</math> and we want <math>\angle{ABC}=3x=\frac{540}{7}</math> to get an answer of <math>\boxed{547}</math>. | ||
+ | ==Solution 3 (Official MAA)== | ||
+ | Let <math>x = \angle ABC = \angle ACB</math>. Because <math>\triangle BCD</math> is isosceles, <math>\angle CBD = 180^\circ - 2x</math>. Then | ||
+ | <cmath>\angle DBE = x - \angle CBD = x - (180^\circ - 2x) = 3x - 180^\circ\!.</cmath>Because <math>\triangle EDA</math> and <math>\triangle DBE</math> are also isosceles, | ||
+ | <cmath>\angle BAC =\frac12(\angle EAD + \angle ADE) = \frac12(\angle BED)= \frac12(\angle DBE)</cmath> | ||
+ | <cmath>= \frac12 (3x - 180^\circ) = \frac32x-90^\circ\!.</cmath> | ||
+ | Because <math>\triangle ABC</math> is isosceles, <math>\angle BAC</math> is also <math>180^\circ-2x</math>, so <math>\frac32x - 90^\circ = 180^\circ - 2x</math>, and it follows that | ||
+ | <math>\angle ABC = x = \left(\frac{540}7\right)^\circ</math>. The requested sum is <math>540+7 = 547</math>. | ||
+ | <asy> | ||
+ | unitsize(4 cm); | ||
+ | |||
+ | pair A, B, C, D, E; | ||
+ | real a = 180/7; | ||
+ | |||
+ | A = (0,0); | ||
+ | B = dir(180 - a/2); | ||
+ | C = dir(180 + a/2); | ||
+ | D = extension(B, B + dir(270 + a), A, C); | ||
+ | E = extension(D, D + dir(90 - 2*a), A, B); | ||
+ | |||
+ | draw(A--B--C--cycle); | ||
+ | draw(B--D--E); | ||
+ | |||
+ | label("$A$", A, dir(0)); | ||
+ | label("$B$", B, NW); | ||
+ | label("$C$", C, SW); | ||
+ | label("$D$", D, S); | ||
+ | label("$E$", E, N); | ||
+ | </asy> | ||
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_25 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_25 | ||
Line 44: | Line 74: | ||
https://youtu.be/4XkA0DwuqYk | https://youtu.be/4XkA0DwuqYk | ||
+ | |||
+ | ==Solution 4 (writing equations)== | ||
+ | graph soon | ||
+ | |||
+ | We write equations based on the triangle sum of angles theorem. There are angles that do not need variables as the less variables the better. | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \angle A &= y \\ | ||
+ | \angle B &= x+z \\ | ||
+ | \angle C &= \frac{180-x}{2} \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Then, using triangle sum of angles theorem, we find that | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \angle A + \angle B + \angle C = x+y+z+\frac{180-x}{2}=180 \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Now we just need to find the variables. | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | (180-2y)+z = 180& \\ | ||
+ | (180-2z)+y+\frac{180-x}{2} = 180& \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Notice how all the equations equal 180. We can use this to write | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | (180-2y)+z = (180-2z)+y+\frac{180-x}{2}=x+y+z+\frac{180-x}{2} \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Simplifying, we get | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | (180-2y)+z=(180-2z)+y+\frac{180-x}{2} \\ | ||
+ | 360-4y+2z=360-4z+2y+180-x \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | 6z=6y+180-x \\ | ||
+ | x=6y-6z+180 \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | (180-2y)+z=6y-6z+180+y+z+\frac{180-(6y-6+180)}{2} \\ | ||
+ | 360-4y+2z=12y-12z+360+2y+2z+180-6y+6z-180 \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | 6z=12y& \\ | ||
+ | z=2y& \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Theres more. We are at a dead end right now because we forgot that the problem states that the triangle is isosceles. With this, we can write the equation | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \frac{180-x}{2}=x+z \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Substituting <math>z</math> with <math>2y</math>, we get | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \frac{180-x}{2}=x+2y \\ | ||
+ | 180-x=2x+4y \\ | ||
+ | \end{align*}</cmath> | ||
+ | <cmath>\begin{align*} | ||
+ | 180-(6y-6z+180)=2(6y-6z+180)+4y& \\ | ||
+ | 180-6y+12y-180=12y-24y+360+4y& \\ | ||
+ | \end{align*}</cmath> | ||
+ | <cmath>\begin{align*} | ||
+ | 6y=-8y+360& \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | With this, we get | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | y=\frac{180}{7} \\ | ||
+ | x=\frac{180}{7} \\ | ||
+ | z=\frac{360}{7} \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | And a final answer of <math>\frac{180}{7}+\frac{360}{7} = \frac{540}{7} = \boxed{547}</math>. | ||
+ | |||
+ | ~[[OrenSH|orenbad]] | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/O_o_-yjGrOU?t=333 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video solution== | ||
+ | |||
+ | https://youtu.be/IH7yM3L5xjA | ||
+ | |||
+ | https://youtu.be/mgRNqSDCvgM ~yofro | ||
+ | |||
+ | ==Solution without words== | ||
+ | [[File:2020 AIME I 1.png|450px|left]] | ||
+ | [[File:2020 AIME I 1a.png|460px|right]] | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=I|before=First Problem|num-a=2}} | {{AIME box|year=2020|n=I|before=First Problem|num-a=2}} | ||
+ | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:01, 24 January 2024
Contents
Problem
In with
point
lies strictly between
and
on side
and point
lies strictly between
and
on side
such that
The degree measure of
is
where
and
are relatively prime positive integers. Find
Solution 1
If we set to
, we can find all other angles through these two properties:
1. Angles in a triangle sum to
.
2. The base angles of an isosceles triangle are congruent.
Now we angle chase. ,
,
,
,
,
. Since
as given by the problem,
, so
. Therefore,
, and our desired angle is
for an answer of
.
See here for a video solution: https://youtu.be/4e8Hk04Ax_E
Solution 2
Let be
in degrees.
.
By Exterior Angle Theorem on triangle
,
.
By Exterior Angle Theorem on triangle
,
.
This tells us
and
.
Thus
and we want
to get an answer of
.
Solution 3 (Official MAA)
Let . Because
is isosceles,
. Then
Because
and
are also isosceles,
Because
is isosceles,
is also
, so
, and it follows that
. The requested sum is
.
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_25 (Almost Mirrored)
See here for a video solution:
Solution 4 (writing equations)
graph soon
We write equations based on the triangle sum of angles theorem. There are angles that do not need variables as the less variables the better.
Then, using triangle sum of angles theorem, we find that
Now we just need to find the variables.
Notice how all the equations equal 180. We can use this to write
Simplifying, we get
Theres more. We are at a dead end right now because we forgot that the problem states that the triangle is isosceles. With this, we can write the equation
Substituting with
, we get
With this, we get
And a final answer of .
Video Solution by OmegaLearn
https://youtu.be/O_o_-yjGrOU?t=333
~ pi_is_3.14
Video solution
https://youtu.be/mgRNqSDCvgM ~yofro
Solution without words
vladimir.shelomovskii@gmail.com, vvsss
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.