Difference between revisions of "2008 AMC 10B Problems/Problem 24"
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==Solution 1== | ==Solution 1== | ||
This solution requires the use of cyclic quadrilateral properties but could be a bit time-consuming during the contest. | This solution requires the use of cyclic quadrilateral properties but could be a bit time-consuming during the contest. | ||
− | To start off, draw a diagram like in solution one and label the points. Now draw <math>\overline{AC}</math> and <math>\overline{BD}</math> and call their intersection point <math>Y</math>. Note that triangle <math>BCD</math> is an isosceles triangle so angles <math>CDB</math> and <math>CBD</math> are each <math>5</math> degrees. Since <math>AB</math> equals <math>BC</math>, angle <math>BAC</math> equals <math>55</math> degrees, thus making angle <math>AYB</math> equal to <math>60</math> degrees. We can also find out that angle <math>CYB</math> equals <math>120</math> degrees. Extend point <math>C</math> such that it lies on the same level of segment <math>AB</math>. Call this point <math>E</math>. Since angle <math>BEC</math> plus angle <math>CYB</math> equals <math>180</math> degrees, quadrilateral <math>YCEB</math> is a cyclic quadrilateral. Next, draw a line from point <math>Y</math> to point <math>E</math>. Since angle <math>YBC</math> and angle <math>YEC</math> point to the same arc, angle <math>YEC</math> is equal to <math>5 | + | To start off, draw a diagram like in solution one and label the points. Now draw <math>\overline{AC}</math> and <math>\overline{BD}</math> and call their intersection point <math>Y</math>. Note that triangle <math>BCD</math> is an isosceles triangle so angles <math>CDB</math> and <math>CBD</math> are each <math>5</math> degrees. Since <math>AB</math> equals <math>BC</math>, angle <math>BAC</math> equals <math>55</math> degrees, thus making angle <math>AYB</math> equal to <math>60</math> degrees. We can also find out that angle <math>CYB</math> equals <math>120</math> degrees. Extend point <math>C</math> such that it lies on the same level of segment <math>AB</math>. Call this point <math>E</math>. Since angle <math>BEC</math> plus angle <math>CYB</math> equals <math>180</math> degrees, quadrilateral <math>YCEB</math> is a cyclic quadrilateral. Next, draw a line from point <math>Y</math> to point <math>E</math>. Since angle <math>YBC</math> and angle <math>YEC</math> point to the same arc, angle <math>YEC</math> is equal to <math>5</math> degrees. Since <math>EYD</math> is an isosceles triangle (based on angle properties) and <math>YAE</math> is also an isosceles triangle, we can find that <math>YAD</math> is also an isosceles triangle. Thus, each of the other angles is <math>\frac{180-120}{2}=30</math> degrees. Finally, we have angle <math>BAD</math> equals <math>30+55=\boxed{85}</math> degrees. |
==Solution 2== | ==Solution 2== |
Revision as of 13:53, 7 June 2021
Contents
Problem
Quadrilateral has
, angle
and angle
. What is the measure of angle
?
Solution 1
This solution requires the use of cyclic quadrilateral properties but could be a bit time-consuming during the contest.
To start off, draw a diagram like in solution one and label the points. Now draw and
and call their intersection point
. Note that triangle
is an isosceles triangle so angles
and
are each
degrees. Since
equals
, angle
equals
degrees, thus making angle
equal to
degrees. We can also find out that angle
equals
degrees. Extend point
such that it lies on the same level of segment
. Call this point
. Since angle
plus angle
equals
degrees, quadrilateral
is a cyclic quadrilateral. Next, draw a line from point
to point
. Since angle
and angle
point to the same arc, angle
is equal to
degrees. Since
is an isosceles triangle (based on angle properties) and
is also an isosceles triangle, we can find that
is also an isosceles triangle. Thus, each of the other angles is
degrees. Finally, we have angle
equals
degrees.
Solution 2
First, connect the diagonal , then, draw line
such that it is congruent to
and is parallel to
. Because triangle
is isosceles and angle
is
, the angles
and
are both
. Because angle
is
, we get angle
is
. Next, noticing parallel lines
and
and transversal
, we see that angle
is also
, and subtracting off angle
gives that angle
is
.
Now, because we drew , triangle
is equilateral. We can also conclude that
meaning that triangle
is isosceles, and angles
and
are equal.
Finally, we can set up our equation. Denote angle as
. Then, because
is a parallelogram, the angle
is also
. Then,
is
. Again because
is a parallelogram, angle
is
. Subtracting angle
gives that angle
equals
. Because angle
equals angle
, we get
, solving into
.
Side note: this solution was inspired by some basic angle chasing and finding some 60 degree angles, which made me want to create equilateral triangles.
~Someonenumber011
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.