Difference between revisions of "2008 AMC 10B Problems/Problem 24"
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<math>\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95</math> | <math>\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95</math> | ||
− | + | ==Solution 1== | |
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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 the <math>\overline{AC}</math> and <math>\overline{BD}</math> and call this 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> had to equal <math>55</math> degrees, thus making angle <math>AYB</math> equal to <math>60</math> degrees. We can also find out that angle CYB 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 degrees</math>. Since <math> | + | To start off, draw a diagram like in solution one and label the points. Now draw the <math>\overline{AC}</math> and <math>\overline{BD}</math> and call this 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> had to equal <math>55</math> degrees, thus making angle <math>AYB</math> equal to <math>60</math> degrees. We can also find out that angle CYB 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 degrees</math>. 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 | + | ==Solution 2== |
First, connect the diagonal <math>DB</math>, then, draw line <math>DE</math> such that it is congruent to <math>DC</math> and is parallel to <math>AB</math>. Because triangle <math>DCB</math> is isosceles and angle <math>DCB</math> is <math>170^\circ</math>, the angles <math>CDB</math> and <math>CBD</math> are both <math>\frac{180-170}{2} = 5^\circ</math>. Because angle <math>ABC</math> is <math>70^\circ</math>, we get angle <math>ABD</math> is <math>65^\circ</math>. Next, noticing parallel lines <math>AB</math> and <math>DE</math> and transversal <math>DB</math>, we see that angle <math>BDE</math> is also <math>65^\circ</math>, and subtracting off angle <math>CDB</math> gives that angle <math>EDC</math> is <math>60^\circ</math>. | First, connect the diagonal <math>DB</math>, then, draw line <math>DE</math> such that it is congruent to <math>DC</math> and is parallel to <math>AB</math>. Because triangle <math>DCB</math> is isosceles and angle <math>DCB</math> is <math>170^\circ</math>, the angles <math>CDB</math> and <math>CBD</math> are both <math>\frac{180-170}{2} = 5^\circ</math>. Because angle <math>ABC</math> is <math>70^\circ</math>, we get angle <math>ABD</math> is <math>65^\circ</math>. Next, noticing parallel lines <math>AB</math> and <math>DE</math> and transversal <math>DB</math>, we see that angle <math>BDE</math> is also <math>65^\circ</math>, and subtracting off angle <math>CDB</math> gives that angle <math>EDC</math> is <math>60^\circ</math>. | ||
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label("$65^\circ$",B + .75*dir(180-32.5)); | label("$65^\circ$",B + .75*dir(180-32.5)); | ||
label("$x^\circ$",A + .5*dir(42.5)); | label("$x^\circ$",A + .5*dir(42.5)); | ||
− | label("$5^\circ$",D + 2.5*dir(360-60-2 | + | label("$5^\circ$",D + 2.5*dir(360-60-2.5)); |
label("$60^\circ$",D + .75*dir(360-30)); | label("$60^\circ$",D + .75*dir(360-30)); | ||
label("$60^\circ$",E + .5*dir(360-150)); | label("$60^\circ$",E + .5*dir(360-150)); | ||
− | label("$5^\circ$",B + 2.5*dir(180-65-2 | + | label("$5^\circ$",B + 2.5*dir(180-65-2.5)); |
</asy> | </asy> | ||
Latest revision as of 18:00, 1 January 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 the and and call this intersection point . Note that triangle is an isosceles triangle so angles and are each degrees. Since equals , angle had to equal degrees, thus making angle equal to degrees. We can also find out that angle CYB 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 . 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 |
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