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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.
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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.
+
 
 +
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 <math>\overline{CD}</math> and <math>\overline{AB}</math> and let their intersection be <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 14:26, 5 August 2021

Problem

Quadrilateral $ABCD$ has $AB = BC = CD$, angle $ABC = 70$ and angle $BCD = 170$. What is the measure of angle $BAD$?

$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$

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 $\overline{AC}$ and $\overline{BD}$ and call their intersection point $Y$. Note that triangle $BCD$ is an isosceles triangle so angles $CDB$ and $CBD$ are each $5$ degrees. Since $AB$ equals $BC$, angle $BAC$ equals $55$ degrees, thus making angle $AYB$ equal to $60$ degrees. We can also find out that angle $CYB$ equals $120$ degrees.

Extend $\overline{CD}$ and $\overline{AB}$ and let their intersection be $E$. Since angle $BEC$ plus angle $CYB$ equals $180$ degrees, quadrilateral $YCEB$ is a cyclic quadrilateral.

Next, draw a line from point $Y$ to point $E$. Since angle $YBC$ and angle $YEC$ point to the same arc, angle $YEC$ is equal to $5$ degrees. Since $EYD$ is an isosceles triangle (based on angle properties) and $YAE$ is also an isosceles triangle, we can find that $YAD$ is also an isosceles triangle. Thus, each of the other angles is $\frac{180-120}{2}=30$ degrees. Finally, we have angle $BAD$ equals $30+55=\boxed{85}$ degrees.

Solution 2

First, connect the diagonal $DB$, then, draw line $DE$ such that it is congruent to $DC$ and is parallel to $AB$. Because triangle $DCB$ is isosceles and angle $DCB$ is $170^\circ$, the angles $CDB$ and $CBD$ are both $\frac{180-170}{2} = 5^\circ$. Because angle $ABC$ is $70^\circ$, we get angle $ABD$ is $65^\circ$. Next, noticing parallel lines $AB$ and $DE$ and transversal $DB$, we see that angle $BDE$ is also $65^\circ$, and subtracting off angle $CDB$ gives that angle $EDC$ is $60^\circ$.

Now, because we drew $ED = DC$, triangle $DEC$ is equilateral. We can also conclude that $EC=DC=CB$ meaning that triangle $ECB$ is isosceles, and angles $CBE$ and $CEB$ are equal.

Finally, we can set up our equation. Denote angle $BAD$ as $x^\circ$. Then, because $ABED$ is a parallelogram, the angle $DEB$ is also $x^\circ$. Then, $CEB$ is $(x-60)^\circ$. Again because $ABED$ is a parallelogram, angle $ABE$ is $(180-x)^\circ$. Subtracting angle $ABC$ gives that angle $CBE$ equals $(110-x)^\circ$. Because angle $CBE$ equals angle $CEB$, we get \[x-60=110-x\], solving into $x=\boxed{85^\circ}$.


[asy] unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120), E=D+a*dir(0); draw(A--B--C--D--cycle); draw(E--C); draw(B--D); draw(B--E); draw(D--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,SE); label("$D$",D,N); label("$E$",E,NE); label("$60^\circ$",C + .75*dir(360-65-115-55-30)); label("$65^\circ$",B + .75*dir(180-32.5)); label("$x^\circ$",A + .5*dir(42.5)); label("$5^\circ$",D + 2.5*dir(360-60-2.5)); label("$60^\circ$",D + .75*dir(360-30)); label("$60^\circ$",E + .5*dir(360-150)); label("$5^\circ$",B + 2.5*dir(180-65-2.5)); [/asy]

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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