Difference between revisions of "2008 AMC 10B Problems/Problem 24"

(Solution)
(Solution 6)
(30 intermediate revisions by 14 users not shown)
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Also note that for similar reasons the triangles <math>CBP</math> and <math>CDP</math> are congruent.
 
Also note that for similar reasons the triangles <math>CBP</math> and <math>CDP</math> are congruent.
  
Now we can compute their inner angles. <math>BP</math> is the bisector of the angle <math>ABC</math>, hence <math>\angle ABP = \angle CBP = 35^\circ</math>, and thus also <math>\angle CDP = 35^\circ</math>. <math>CP</math> is the bisector of the angle <math>BCD</math>, hence <math>\angle BCP = \angle DCP = 85^\circ</math>, and thus also <math>\angle BAP = 85^\circ</math>.
+
Now we can compute their inner angles. <math>BP</math> is the bisector of the angle <math>ABC</math>, hence <math>\angle ABP = \angle CBP = 35^\circ</math>, and thus also <math>\angle CDP = 35^\circ</math>. (Faster Solution picks up here) <math>CP</math> is the bisector of the angle <math>BCD</math>, hence <math>\angle BCP = \angle DCP = 85^\circ</math>, and thus also <math>\angle BAP = 85^\circ</math>.
  
It follows that <math>\angle APB = \angle BPC = \angle CPD = 180^\circ - 35^\circ - 85^\circ = 60^\circ</math>. Thus the angle <math>APB</math> has <math>180^\circ</math>, and hence <math>P</math> does indeed lie on <math>AD</math>. Then obviously <math>\angle BAD = \angle BAP = \boxed{ 85^\circ }</math>.
+
It follows that <math>\angle APB = \angle BPC = \angle CPD = 180^\circ - 35^\circ - 85^\circ = 60^\circ</math>. Thus the angle <math>APD</math> has <math>180^\circ</math>, and hence <math>P</math> does indeed lie on <math>AD</math>. Then obviously <math>\angle BAD = \angle BAP = \boxed{ 85^\circ }</math>.
  
 
<asy>
 
<asy>
Line 38: Line 38:
 
label("$85^\circ$",C + .5*dir(120+85+42.5));
 
label("$85^\circ$",C + .5*dir(120+85+42.5));
 
</asy>
 
</asy>
 +
 +
Faster Solution: Because we now know three angles, we can subtract to get <math>360 - 35 - 85 - 85 - 35 - 35</math>, or <math>\boxed{85}</math>.
 +
 +
 +
Even Faster Solution: Above, we proved that P falls on line AD, and also <math>\triangle ABP = \triangle CBP</math>, by <math>SAS</math>, hence we have <math>\angle BCP=\angle BAP</math>, which is the angle bisector of <math>\angle BCD</math> which is <math>\dfrac{170}{2}=85</math>. Hence we have <math>\angle BCP=\angle BAP=\angle BAD= 85^\circ</math>
  
 
=== Solution 2 ===
 
=== Solution 2 ===
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label("$E$",P,W);
 
label("$E$",P,W);
 
</asy>
 
</asy>
 +
 +
=== Solution 4 ===
 +
Start off with the same diagram as solution 1. Now draw <math>\overline{CA}</math> which creates isosceles <math>\triangle CAB</math>. We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that it's is <math>\boxed{85}.</math>
 +
 +
==Solution 5==
 +
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>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.
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 +
==Solution 6==
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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>.
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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.
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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>.
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 +
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<asy>
 +
unitsize(1cm);
 +
defaultpen(.8);
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real a=4;
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pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120), E=D+a*dir(0);
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draw(A--B--C--D--cycle);
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draw(E--C);
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draw(B--D);
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draw(B--E);
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draw(D--E);
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label("$A$",A,SW);
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label("$B$",B,SE);
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label("$C$",C,SE);
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label("$D$",D,N);
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label("$E$",E,NE);
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label("$60^\circ$",C + .75*dir(360-65-115-55-30));
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label("$65^\circ$",B + .75*dir(180-32.5));
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label("$x^\circ$",A + .5*dir(42.5));
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label("$5^\circ$",D + 2.5*dir(360-60-2.5));
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label("$60^\circ$",D + .75*dir(360-30));
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label("$60^\circ$",E + .5*dir(360-150));
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label("$5^\circ$",B + 2.5*dir(180-65-2.5));
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</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==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=23|num-a=25}}
 
{{AMC10 box|year=2008|ab=B|num-b=23|num-a=25}}
 +
{{MAA Notice}}

Revision as of 11:59, 24 May 2020

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

Solution 1

Draw the angle bisectors of the angles $ABC$ and $BCD$. These two bisectors obviously intersect. Let their intersection be $P$. We will now prove that $P$ lies on the segment $AD$.

Note that the triangles $ABP$ and $CBP$ are congruent, as they share the side $BP$, and we have $AB=BC$ and $\angle ABP = \angle CBP$.

Also note that for similar reasons the triangles $CBP$ and $CDP$ are congruent.

Now we can compute their inner angles. $BP$ is the bisector of the angle $ABC$, hence $\angle ABP = \angle CBP = 35^\circ$, and thus also $\angle CDP = 35^\circ$. (Faster Solution picks up here) $CP$ is the bisector of the angle $BCD$, hence $\angle BCP = \angle DCP = 85^\circ$, and thus also $\angle BAP = 85^\circ$.

It follows that $\angle APB = \angle BPC = \angle CPD = 180^\circ - 35^\circ - 85^\circ = 60^\circ$. Thus the angle $APD$ has $180^\circ$, and hence $P$ does indeed lie on $AD$. Then obviously $\angle BAD = \angle BAP = \boxed{ 85^\circ }$.

[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); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,N); label("$P$",P,W);  label("$35^\circ$",B + dir(180-17.5)); label("$35^\circ$",B + dir(180-35-17.5));  label("$85^\circ$",C + .5*dir(120+42.5)); label("$85^\circ$",C + .5*dir(120+85+42.5)); [/asy]

Faster Solution: Because we now know three angles, we can subtract to get $360 - 35 - 85 - 85 - 35 - 35$, or $\boxed{85}$.


Even Faster Solution: Above, we proved that P falls on line AD, and also $\triangle ABP = \triangle CBP$, by $SAS$, hence we have $\angle BCP=\angle BAP$, which is the angle bisector of $\angle BCD$ which is $\dfrac{170}{2}=85$. Hence we have $\angle BCP=\angle BAP=\angle BAD= 85^\circ$

Solution 2

Draw the diagonals $\overline{BD}$ and $\overline{AC}$, and suppose that they intersect at $E$. Then, $\triangle ABC$ and $\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\angle BAC = 55^{\circ}$, $\angle CBD = 5^{\circ}$, and $\angle BEA = 180 - \angle EBA - \angle BAE = 60^{\circ}$. Draw $E'$ such that $EE'B = 60^{\circ}$ and so that $E'$ is on $\overline{AE}$, and draw $E''$ such that $\angle EE''C = 60^{\circ}$ and $E''$ is on $\overline{DE}$. It follows that $\triangle BEE'$ and $\triangle CEE''$ are both equilateral. Also, it is easy to see that $\triangle BEC \cong \triangle DE''C$ and $\triangle BCE \cong \triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\triangle ADE$ is isosceles. Since $\angle AED = 120^{\circ}$, then $\angle DAC = \frac{180 - 120}{2} = 30^{\circ}$, and $\angle BAD = 30 + 55 = 85^{\circ}$. [asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947;  pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0);  filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94));  dot((0,0),ds); label("$A$",(-0.096,0.005),NE*lsf); dot((1,0),ds); label("$B$",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label("$C$",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label("$D$",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label("$E$",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label("$E'$",(0.1,0.23),NE*lsf); label("$60^\circ$",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label("$E''$",(0.423,0.957),NE*lsf); label("$60^\circ$",(0.761,0.886),NE*lsf,qqwuqq);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  [/asy]

Solution 3

Again, draw the diagonals $\overline{BD}$ and $\overline{AC}$, and suppose that they intersect at $E$. We find by angle chasing the same way as in solution 2 that $m\angle ABE = 65^\circ$ and $m\angle DCE = 115^\circ$. Applying the Law of Sines to $\triangle AEB$ and $\triangle EDC$, it follows that $DE = \frac{2\sin 115^\circ}{\sin \angle DEC} = \frac{2\sin 65^\circ}{\sin \angle AEB} = EA$, so $\triangle AED$ is isosceles. We finish as we did in solution 2.

[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); draw(A--B--C--D--cycle); pair P=intersectionpoint(B--D,C--A); draw(A--C); draw(B--D); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,N); label("$E$",P,W); [/asy]

Solution 4

Start off with the same diagram as solution 1. Now draw $\overline{CA}$ which creates isosceles $\triangle CAB$. We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that it's is $\boxed{85}.$

Solution 5

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 $\overline{AC}$ and $\overline{BD}$ and call this 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 $BAC$ had to equal $55$ degrees, thus making angle $AYB$ equal to $60$ degrees. We can also find out that angle CYB equals $120$ degrees. Extend point $C$ such that it lies on the same level of segment $AB$. Call this point $E$. 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 $EBD$ 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 6

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 (ProblemsAnswer KeyResources)
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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