https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Omer.dokan1&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T00:31:01ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10B_Problems/Problem_24&diff=1388182008 AMC 10B Problems/Problem 242020-11-30T03:06:08Z<p>Omer.dokan1: /* Solution */</p>
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<div>==Problem==<br />
Quadrilateral <math>ABCD</math> has <math>AB = BC = CD</math>, angle <math>ABC = 70</math> and angle <math>BCD = 170</math>. What is the measure of angle <math>BAD</math>?<br />
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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><br />
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==Solution 5==<br />
This solution requires the use of cyclic quadrilateral properties but could be a bit time-consuming during the contest.<br />
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>EBD</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.<br />
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==Solution 6==<br />
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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>.<br />
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Now, because we drew <math>ED = DC</math>, triangle <math>DEC</math> is equilateral. We can also conclude that <math>EC=DC=CB</math> meaning that triangle <math>ECB</math> is isosceles, and angles <math>CBE</math> and <math>CEB</math> are equal.<br />
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Finally, we can set up our equation. Denote angle <math>BAD</math> as <math>x^\circ</math>. Then, because <math>ABED</math> is a parallelogram, the angle <math>DEB</math> is also <math>x^\circ</math>. Then, <math>CEB</math> is <math>(x-60)^\circ</math>. Again because <math>ABED</math> is a parallelogram, angle <math>ABE</math> is <math>(180-x)^\circ</math>. Subtracting angle <math>ABC</math> gives that angle <math>CBE</math> equals <math>(110-x)^\circ</math>. Because angle <math>CBE</math> equals angle <math>CEB</math>, we get <cmath>x-60=110-x</cmath>, solving into <math>x=\boxed{85^\circ}</math>.<br />
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<asy><br />
unitsize(1cm);<br />
defaultpen(.8);<br />
real a=4;<br />
pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120), E=D+a*dir(0);<br />
draw(A--B--C--D--cycle);<br />
draw(E--C);<br />
draw(B--D);<br />
draw(B--E);<br />
draw(D--E);<br />
label("$A$",A,SW);<br />
label("$B$",B,SE);<br />
label("$C$",C,SE);<br />
label("$D$",D,N);<br />
label("$E$",E,NE);<br />
label("$60^\circ$",C + .75*dir(360-65-115-55-30));<br />
label("$65^\circ$",B + .75*dir(180-32.5));<br />
label("$x^\circ$",A + .5*dir(42.5));<br />
label("$5^\circ$",D + 2.5*dir(360-60-2.5));<br />
label("$60^\circ$",D + .75*dir(360-30));<br />
label("$60^\circ$",E + .5*dir(360-150));<br />
label("$5^\circ$",B + 2.5*dir(180-65-2.5));<br />
</asy><br />
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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.<br />
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~Someonenumber011<br />
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==See also==<br />
{{AMC10 box|year=2008|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Omer.dokan1