https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=LOTRFan123&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T08:48:59ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2010_AIME_II_Problems/Problem_14&diff=770292010 AIME II Problems/Problem 142016-02-25T04:54:23Z<p>LOTRFan123: /* Solution */</p>
<hr />
<div>== Problem ==<br />
[[Triangle]] <math>ABC</math> with [[right angle]] at <math>C</math>, <math>\angle BAC < 45^\circ</math> and <math>AB = 4</math>. Point <math>P</math> on <math>\overline{AB}</math> is chosen such that <math>\angle APC = 2\angle ACP</math> and <math>CP = 1</math>. The ratio <math>\frac{AP}{BP}</math> can be represented in the form <math>p + q\sqrt{r}</math>, where <math>p</math>, <math>q</math>, <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime. Find <math>p+q+r</math>.<br />
<br />
== Solution ==<br />
Let <math>O</math> be the [[circumcenter]] of <math>ABC</math> and let the intersection of <math>CP</math> with the [[circumcircle]] be <math>D</math>. It now follows that <math>\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}</math>. Hence <math>ODP</math> is isosceles and <math>OD = DP = 2</math>. <br />
<br />
Denote <math>E</math> the projection of <math>O</math> onto <math>CD</math>. Now <math>CD = CP + DP = 3</math>. By the [[pythagorean theorem]], <math>OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}</math>. Now note that <math>EP = \frac {1}{2}</math>. By the pythagorean theorem, <math>OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}</math>. Hence it now follows that,<br />
<br />
<cmath>\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}</cmath><br />
<br />
This gives that the answer is <math>\boxed{007}</math>.<br />
<br />
An alternate finish for this problem would be to use Power of a Point on <math>BA</math> and <math>CD</math>. By Power of a Point Theorem, <math>CP\cdot PD=1\cdot 2=BP\cdot PA</math>. Since <math>BP+PA=4</math>, we can solve for <math>BP</math> and <math>PA</math>, giving the same values and answers as above. <br />
<br />
<center><asy> /* geogebra conversion, see azjps userscripts.org/scripts/show/72997 */<br />
import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(250); real lsf = 0.5; /* changes label-to-point distance */<br />
pen xdxdff = rgb(0.49,0.49,1); pen qqwuqq = rgb(0,0.39,0); pen fftttt = rgb(1,0.2,0.2);<br />
<br />
/* segments and figures */<br />
draw((0.2,0.81)--(0.33,0.78)--(0.36,0.9)--(0.23,0.94)--cycle,qqwuqq); draw((0.81,-0.59)--(0.93,-0.54)--(0.89,-0.42)--(0.76,-0.47)--cycle,qqwuqq); draw(circle((2,0),2)); draw((0,0)--(0.23,0.94),linewidth(1.6pt)); draw((0.23,0.94)--(4,0),linewidth(1.6pt)); draw((0,0)--(4,0),linewidth(1.6pt)); draw((0.23,(+0.55-0.94*0.23)/0.35)--(4.67,(+0.55-0.94*4.67)/0.35));<br />
<br />
/* points and labels */<br />
label("$1$", (0.26,0.42), SE*lsf); draw((1.29,-1.87)--(2,0)); label("$2$", (2.91,-0.11), SE*lsf); label("$2$", (1.78,-0.82), SE*lsf); pair parametricplot0_cus(real t){<br />
return (0.28*cos(t)+0.23,0.28*sin(t)+0.94);<br />
}<br />
draw(graph(parametricplot0_cus,-1.209429202888189,-0.24334747753738661)--(0.23,0.94)--cycle,fftttt); pair parametricplot1_cus(real t){<br />
return (0.28*cos(t)+0.59,0.28*sin(t)+0);<br />
}<br />
draw(graph(parametricplot1_cus,0.0,1.9321634507016043)--(0.59,0)--cycle,fftttt); label("$\theta$", (0.42,0.77), SE*lsf); label("$2\theta$", (0.88,0.38), SE*lsf); draw((2,0)--(0.76,-0.47)); pair parametricplot2_cus(real t){<br />
return (0.28*cos(t)+2,0.28*sin(t)+0);<br />
}<br />
draw(graph(parametricplot2_cus,-1.9321634507016048,0.0)--(2,0)--cycle,fftttt); label("$2\theta$", (2.18,-0.3), SE*lsf); dot((0,0)); label("$B$", (-0.21,-0.2),NE*lsf); dot((4,0)); label("$A$", (4.03,0.06),NE*lsf); dot((2,0)); label("$O$", (2.04,0.06),NE*lsf); dot((0.59,0)); label("$P$", (0.28,-0.27),NE*lsf); dot((0.23,0.94)); label("$C$", (0.07,1.02),NE*lsf); dot((1.29,-1.87)); label("$D$", (1.03,-2.12),NE*lsf); dot((0.76,-0.47)); label("$E$", (0.56,-0.79),NE*lsf); clip((-0.92,-2.46)--(-0.92,2.26)--(4.67,2.26)--(4.67,-2.46)--cycle);<br />
</asy></center><br />
<br />
==Solution 2==<br />
Let <math>AC=b</math>, <math>BC=a</math> by convention. Also, Let <math>AP=x</math> and <math>BP=y</math>. Finally, let <math> \angle ACP=\theta</math> and <math> \angle APC=2\theta</math>. <br />
<br />
We are then looking for <math> \frac{AP}{BP}=\frac{x}{y}</math><br />
<br />
Now, by arc interceptions and angle chasing we find that <math> \triangle BPD \sim \triangle CPA</math>, and that therefore <math> BD=yb.</math> Then, since <math> \angle ABD=\theta</math> (it intercepts the same arc as <math> \angle ACD</math>) and <math> ADB</math> is right, <br />
<br />
<math> \cos\theta=\frac{DB}{AB}=\frac{by}{4}</math>. <br />
<br />
<br />
Using law of sines on <math>APC</math>, we additionally find that <math> \frac{b}{\sin 2\theta}=\frac{x}{\sin\theta}.</math> Simplification by the double angle formula <math> \sin 2\theta=2\sin \theta\cos\theta</math> yields <br />
<br />
<math> \cos \theta=\frac{b}{2x}</math>.<br />
<br />
<br />
We equate these expressions for <math> \cos\theta</math> to find that <math> xy=2</math>. Since <math> x+y=AB=4</math>, we have enough information to solve for <math>x</math> and <math>y</math>. We obtain <math> x,y=2 \pm \sqrt{2}</math> <br />
<br />
Since we know <math>x>y</math>, we use <math> \frac{x}{y}=\frac{2+\sqrt{2}}{2-\sqrt{2}}=3+2\sqrt{2}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2010|num-b=13|num-a=15|n=II}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>LOTRFan123https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_20&diff=717341982 AHSME Problems/Problem 202015-08-22T18:15:55Z<p>LOTRFan123: /* Solution */</p>
<hr />
<div>==1982 AHSME Problems/Problem 20==<br />
<br />
==Problem==<br />
<br />
The number of pairs of positive integers <math>(x,y)</math> which satisfy the equation <math>x^2+y^2=x^3</math> is<br />
<br />
<math>\text {(A)} 0 \qquad \text {(B)} 1 \qquad \text {(C)} 2 \qquad \text {(D)} \text{not finite} \qquad \text {(E)} \text{none of these}</math><br />
<br />
==Solution==<br />
<br />
Rearrange the equation to <math>y^2=x^3-x^2=x^2(x-1)</math>. This equation is satisfied whenever <math>x-1</math> is a perfect square. There are infinite possible values of <math>x</math>, and thus the answer is <math>\boxed{D: \text{Not Finite}}</math></div>LOTRFan123https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_20&diff=717331982 AHSME Problems/Problem 202015-08-22T18:15:31Z<p>LOTRFan123: Created page with "==1982 AHSME Problems/Problem 20== ==Problem== The number of pairs of positive integers <math>(x,y)</math> which satisfy the equation <math>x^2+y^2=x^3</math> is <math>\tex..."</p>
<hr />
<div>==1982 AHSME Problems/Problem 20==<br />
<br />
==Problem==<br />
<br />
The number of pairs of positive integers <math>(x,y)</math> which satisfy the equation <math>x^2+y^2=x^3</math> is<br />
<br />
<math>\text {(A)} 0 \qquad \text {(B)} 1 \qquad \text {(C)} 2 \qquad \text {(D)} \text{not finite} \qquad \text {(E)} \text{none of these}</math><br />
<br />
==Solution==<br />
<br />
Rearrange the equation to <math>y^2=x^3-x^2=x^2(x-1)</math>. This equation is satisfied whenever <math>x-1</math> is a perfect square. There are infinite possible values of <math>x</math>, and thus the answer is <math>\boxed{D: Not Finite}</math></div>LOTRFan123https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_14&diff=717321982 AHSME Problems/Problem 142015-08-22T18:04:29Z<p>LOTRFan123: /* Problem 14: */</p>
<hr />
<div>==1982 AHSME Problems/Problem 14==<br />
<br />
==Problem 14:==<br />
<br />
In the adjoining figure, points <math>B</math> and <math>C</math> lie on line segment <math>AD</math>, and <math>AB, BC</math>, and <math>CD</math> are diameters of circle <math>O, N</math>, and <math>P</math>, respectively. Circles <math>O, N</math>, and <math>P</math> all have radius <math>15</math> and the line <math>AG</math> is tangent to circle <math>P</math> at <math>G</math>. If <math>AG</math> intersects circle <math>N</math> at points <math>E</math> and <math>F</math>, then chord <math>EF</math> has length<br />
<br />
[asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("<math>A</math>", A, W); label("<math>B</math>", B, SE); label("<math>C</math>", C, NE); label("<math>D</math>", D, dir(0)); label("<math>P</math>", P, S); label("<math>N</math>", N, S); label("<math>O</math>", O, S); label("<math>E</math>", E, dir(120)); label("<math>F</math>", F, NE); label("<math>G</math>", G, dir(100));[/asy]<br />
<br />
==Solution:==<br />
<br />
Since <math>GP</math> is 15, <math>AP</math> is 75, and <math>\angle{AGP}=90</math>, <math>AG=15\sqrt{24}</math>. <br />
<br />
Now drop an altitude from <math>N</math> to <math>AG</math> at point <math>H</math>. <math>AN=45</math>, and since <math>\triangle{AGP}</math> is similar to <math>\triangle{AHN}</math>. <math>NH=9</math>. <math>NE=NF=15</math> so by Pythagorean Theorem, <math>EH=HF=12</math>. Thus <math>EF=\boxed{24}</math>. <math>\boxed{C}</math></div>LOTRFan123https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_14&diff=717311982 AHSME Problems/Problem 142015-08-22T18:03:24Z<p>LOTRFan123: </p>
<hr />
<div>==1982 AHSME Problems/Problem 14==<br />
<br />
==Problem 14:==<br />
<br />
In the adjoining figure, points <math>B</math> and <math>C</math> lie on line segment <math>AD</math>, and <math>AB, BC</math>, and <math>CD</math> are diameters of circle <math>O, N</math>, and <math>P</math>, respectively. Circles <math>O, N</math>, and <math>P</math> all have radius <math>15</math> and the line <math>AG</math> is tangent to circle <math>P</math> at <math>G</math>. If <math>AG</math> intersects circle <math>N</math> at points <math>E</math> and <math>F</math>, then chord <math>EF</math> has length<br />
<br />
<math>[asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("</math>A<math>", A, W); label("</math>B<math>", B, SE); label("</math>C<math>", C, NE); label("</math>D<math>", D, dir(0)); label("</math>P<math>", P, S); label("</math>N<math>", N, S); label("</math>O<math>", O, S); label("</math>E<math>", E, dir(120)); label("</math>F<math>", F, NE); label("</math>G<math>", G, dir(100));[/asy]</math><br />
<br />
==Solution:==<br />
<br />
Since <math>GP</math> is 15, <math>AP</math> is 75, and <math>\angle{AGP}=90</math>, <math>AG=15\sqrt{24}</math>. <br />
<br />
Now drop an altitude from <math>N</math> to <math>AG</math> at point <math>H</math>. <math>AN=45</math>, and since <math>\triangle{AGP}</math> is similar to <math>\triangle{AHN}</math>. <math>NH=9</math>. <math>NE=NF=15</math> so by Pythagorean Theorem, <math>EH=HF=12</math>. Thus <math>EF=\boxed{24}</math>. <math>\boxed{C}</math></div>LOTRFan123https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_14&diff=717301982 AHSME Problems/Problem 142015-08-22T18:01:59Z<p>LOTRFan123: </p>
<hr />
<div><math>\bold{1982 AHSME Problems/Problem 14}</math><br />
<br />
<math>\bold{Problem 14:}</math><br />
<br />
In the adjoining figure, points <math>B</math> and <math>C</math> lie on line segment <math>AD</math>, and <math>AB, BC</math>, and <math>CD</math> are diameters of circle <math>O, N</math>, and <math>P</math>, respectively. Circles <math>O, N</math>, and <math>P</math> all have radius <math>15</math> and the line <math>AG</math> is tangent to circle <math>P</math> at <math>G</math>. If <math>AG</math> intersects circle <math>N</math> at points <math>E</math> and <math>F</math>, then chord <math>EF</math> has length<br />
<br />
[asy] size(250); defaultpen(fontsize(10)); pair A=origin, O=(1,0), B=(2,0), N=(3,0), C=(4,0), P=(5,0), D=(6,0), G=tangent(A,P,1,2), E=intersectionpoints(A--G, Circle(N,1))[0], F=intersectionpoints(A--G, Circle(N,1))[1]; draw(Circle(O,1)^^Circle(N,1)^^Circle(P,1)^^G--A--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F^^G^^O^^N^^P); label("<math>A</math>", A, W); label("<math>B</math>", B, SE); label("<math>C</math>", C, NE); label("<math>D</math>", D, dir(0)); label("<math>P</math>", P, S); label("<math>N</math>", N, S); label("<math>O</math>", O, S); label("<math>E</math>", E, dir(120)); label("<math>F</math>", F, NE); label("<math>G</math>", G, dir(100));[/asy]<br />
<br />
<math>\bold{Solution:}</math><br />
<br />
Since <math>GP</math> is 15, <math>AP</math> is 75, and <math>\angle{AGP}=90</math>, <math>AG=15\sqrt{24}</math>. <br />
<br />
Now drop an altitude from <math>N</math> to <math>AG</math> at point <math>H</math>. <math>AN=45</math>, and since <math>\triangle{AGP}</math> is similar to <math>\triangle{AHN}</math>. <math>NH=9</math>. <math>NE=NF=15</math> so by Pythagorean Theorem, <math>EH=HF=12</math>. Thus <math>EF=\boxed{24}</math>. <math>\boxed{C}</math></div>LOTRFan123https://artofproblemsolving.com/wiki/index.php?title=1982_AHSME_Problems/Problem_14&diff=717291982 AHSME Problems/Problem 142015-08-22T18:00:21Z<p>LOTRFan123: Created page with "Solution: Since <math>GP</math> is 15, <math>AP</math> is 75, and <math>\angle{AGP}=90</math>, <math>AG=15\sqrt{24}</math>. Now drop an altitude from <math>N</math> to <mat..."</p>
<hr />
<div>Solution:<br />
<br />
Since <math>GP</math> is 15, <math>AP</math> is 75, and <math>\angle{AGP}=90</math>, <math>AG=15\sqrt{24}</math>. <br />
<br />
Now drop an altitude from <math>N</math> to <math>AG</math> at point <math>H</math>. <math>AN=45</math>, and since <math>\triangle{AGP}</math> is similar to <math>\triangle{AHN}</math>. <math>NH=9</math>. <math>NE=NF=15</math> so by Pythagorean Theorem, <math>EH=HF=12</math>. Thus <math>EF=\boxed{24}</math>. <math>\boxed{C}</math></div>LOTRFan123https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_8&diff=685912012 AIME II Problems/Problem 82015-03-09T01:00:42Z<p>LOTRFan123: /* Problem 8 */</p>
<hr />
<div>== Problem 8 ==<br />
The complex numbers <math>z</math> and <math>w</math> satisfy the system <cmath> z + \frac{20i}w = 5+i </cmath><br />
<cmath> w+\frac{12i}z = -4+10i </cmath> Find the smallest possible value of <math>\vert zw\vert^2</math>.<br />
<br />
== Solution ==<br />
Multiplying the two equations together gives us <cmath>zw + 32i - \frac{240}{zw} = -30 + 46i</cmath> and multiplying by <math>zw</math> then gives us a quadratic in <math>zw</math>: <cmath>(zw)^2 + (30-14i)zw - 240 =0.</cmath> Using the quadratic formula, we find the two possible values of <math>zw</math> to be <math>7i-15 \pm \sqrt{(30-14i)^2 + 240}</math> = <math>6+2i,</math> <math>12i - 36.</math> The smallest possible value of <math>\vert zw\vert^2</math> is then obviously <math>6^2 + 2^2 = \boxed{040.}</math><br />
<br />
== See Also ==<br />
{{AIME box|year=2012|n=II|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>LOTRFan123https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_23&diff=583642007 AMC 12B Problems/Problem 232013-12-23T19:47:43Z<p>LOTRFan123: /* Solution 2 */</p>
<hr />
<div>==Problem 23==<br />
How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to <math>3</math> times their perimeters?<br />
<br />
<math>\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12</math><br />
<br />
==Solution==<br />
<math>\fracqwb = 3(a+bet)</math><br />
qwetwqet<br />
qwet<br />
Using Euclid's formula for generating primitive triples:wqe<br />
<math>a = m^2-n^2</math>, <math>b=2we</math>, <math>c=m^2+n^2</math> whwqtere <math>m</math> and <math>n</math> are relatively prime positive integers, exactly one of which being eveet=2kmn<math>, </math>c=k(mqw^2+n^2)<math><br />
wettqwwqetqwet<br />
<br />
</math>n(m-n)k = 6qwe<math><br />
<br />
Now we dqweto some casework.<br />
<br />
For </math>k=1<math>et<br />
</math>n(m-n) = 6<math> wethich has solutions </math>(7,1)<math>, </math>(5,2)<math>, </math>(5,3)<math>, </math>(7,6)<math><br />
ethe conditions of Euclid's formula, the only solutions are </math>(5,2)<math> and </math>(7,6et<math><br />
<br />
For </math>k=2<math><br />
<br />
</math>n(m-n)=3<math> has solutions </math>(4,qwet1)<math>, </math>(4,3)<math>, both of which are valid.<br />
<br />
For </math>k=3<math><br />
<br />
</math>n(m-n)=2<math> has solutions </math>(3,1)<math>, </math>(3,2)<math> of which only </math>(3,2)<math> is valid.<br />
<br />
For </math>k=6<math><br />
<br />
</math>n(m-n)=1<math> has solution </math>(1,2)<math>, which is valid.<br />
<br />
This means that the solutions for </math>(m,n,k)<math> are<br />
<br />
</math>(5,2,1), (7,6,1), (4,1,2), (4,3,2), (3,2,3), (1,2,6)<math><br />
<br />
</math>6<math> solutions </math>\Rightarrow \mathrm{(A)}$<br />
<br />
==Solution 2==<br />
Let <math>a</math> and <math>b</math> be the two legs of the triangle.<br />
<br />
We have <math>\frac{1}{2}ab = 3(a+b+c)</math>.<br />
<br />
Then <math>ab=6\cdot (a+b+\sqrt {a^2 + b^2})</math>.<br />
<br />
We can complete the square under the root, and we get, <math>ab=6\cdot (a+b+\sqrt {(a+b)^2 - 2ab})</math>.<br />
<br />
Let <math>ab=p</math> and <math>a+b=s</math>, we have <math>p=6\cdot (s+ \sqrt {s^2 - 2p})</math>.<br />
<br />
After rearranging, squaring both sides, and simplifying, we have <math>p=12s-72</math>.<br />
<br />
<br />
Putting back <math>a</math> and <math>b</math>, and after factoring using <math>SFFT</math>, we've got <math>(a-12)\cdot (b-12)=72</math>.<br />
<br />
<br />
Factoring 72, we get 6 pairs of <math>a</math> and <math>b</math><br />
<br />
<br />
<math>(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).</math><br />
<br />
<br />
And this gives us <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2007|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>LOTRFan123https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_13&diff=528822013 AMC 12B Problems/Problem 132013-05-29T19:07:07Z<p>LOTRFan123: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The internal angles of quadrilateral <math>ABCD</math> form an arithmetic progression. Triangles <math>ABD</math> and <math>DCB</math> are similar with <math>\angle DBA = \angle DCB</math> and <math>\angle ADB = \angle CBD</math>. Moreover, the angles in each of these two triangles also form an arithemetic progression. In degrees, what is the largest possible sum of the two largest angles of <math>ABCD</math>?<br />
<br />
<math>\textbf{(A)}\ 210 \qquad \textbf{(B)}\ 220 \qquad \textbf{(C)}\ 230 \qquad \textbf{(D)}\ 240 \qquad \textbf{(E)}\ 250</math><br />
<br />
==Solution==<br />
Since the angles of Quadrilateral ABCD form an arithmetic sequence, we can assign each angle with the value a, a+d, a+2d, and a+3d. Also, since these angles form an arithmetic progression, we can reason out that (a)+(a+3d)=(a+d)+(a+2d)=180.<br />
<br />
For the sake of simplicity, lets rename the angles of each similar triangle. Lets call Angle DBA and Angle DCB Angle 1. Also we rename Angle ADB and Angle CBD Angle 2. Finally we rename Angles BAD and BDC Angle 3.<br />
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Now we can rename the four angles of Quadrilateral ABCD as Angle 2, Angle 1 + 2, Angle 3, and Angle 1 + 3.<br />
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As for the similar triangles, whose Angles are equivalent, we can name them y, y+b, and y+2b. Therefore y+y+b+y+2b=180 and y+b=60. Because these 3 angles are each equal to one of the angles we named Angles 1, 2, and 3, we know that one of these three angles is equal to 60 degrees.<br />
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Now we we use trial and error to find out which of these 3 angles has a value of 60. If we substitute 60 degrees into Angle 1. This would cause the angle values of ABCD to be Angle 2, 60+Angle 2, Angle 3, and 60 + Angle 3. Since these four angles add up to 360, then Angle 2 + Angle 3 = 120. If we list them in increasing value, we get Angle 2, Angle 3, 60 + Angle 2, 60+Angle 3. Note that this is the only sequence that works because the common difference between each term cannot equal or exceed 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, Angle 1, 2, and 3, the angles of both similar triangles, also form an arithmetic sequence with 45, 60, and 75, and the largest two angles of the quadrilateral add up to 240 which is an answer choice.<br />
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If we apply the same reasoning to Angles 2 and 3, we would get the sum of the highest two angles as 220, which works but is lower than 240. Therefore, 240 [D] is the correct answer.<br />
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== See also ==<br />
{{AMC12 box|year=2013|ab=B|num-b=12|num-a=14}}<br />
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[[Category:Introductory Geometry Problems]]</div>LOTRFan123https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_13&diff=528812013 AMC 12B Problems/Problem 132013-05-29T19:03:53Z<p>LOTRFan123: /* Solution */</p>
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<div>==Problem==<br />
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The internal angles of quadrilateral <math>ABCD</math> form an arithmetic progression. Triangles <math>ABD</math> and <math>DCB</math> are similar with <math>\angle DBA = \angle DCB</math> and <math>\angle ADB = \angle CBD</math>. Moreover, the angles in each of these two triangles also form an arithemetic progression. In degrees, what is the largest possible sum of the two largest angles of <math>ABCD</math>?<br />
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<math>\textbf{(A)}\ 210 \qquad \textbf{(B)}\ 220 \qquad \textbf{(C)}\ 230 \qquad \textbf{(D)}\ 240 \qquad \textbf{(E)}\ 250</math><br />
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==Solution==<br />
Since the angles of Quadrilateral ABCD form an arithmetic sequence, we can assign each angle with the value a, a+d, a+2d, and a+3d. Also, since these angles form an arithmetic progression, we can reason out that (a)+(a+3d)=(a+d)+(a+2d)=180.<br />
<br />
For the sake of simplicity, lets rename the angles of each similar triangle. Lets call Angle DBA and Angle DCB Angle 1. Also we rename Angle ADB and Angle CBD Angle 2. Finally we rename Angles BAD and BDC Angle 3.<br />
<br />
Now we can rename the four angles of Quadrilateral ABCD as Angle 2, Angle 1 + 2, Angle 3, and Angle 1 + 3.<br />
<br />
As for the similar triangles, whose Angles are equivalent, we can name them y, y+b, and y+2b. Therefore y+y+b+y+2b=180 and y+b=60. Because these 3 angles are each equal to one of the angles we named Angles 1, 2, and 3, we know that one of these three angles is equal to 60 degrees.<br />
<br />
Now we we use trial and error to find out which of these 3 angles has a value of 60. If we substitute 60 degrees into Angle 1. This would cause the angle values of ABCD to be Angle 2, 60+Angle 2, Angle 3, and 60 + Angle 3. Since these four angles add up to 360, then Angle 2 + Angle 3 = 120. If we list them in increasing value, we get Angle 2, Angle 3, 60 + Angle 2, 60+Angle 3. Note that this is the only sequence that works because the common difference between each term cannot equal or exceed 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, Angle 1, 2, and 3 also form an arithmetic sequence with 45, 60, and 75, and the largest two angles of the quadrilateral add up to 240 which is an answer choice.<br />
<br />
If we apply the same reasoning to Angles 2 and 3, we would get the sum of the highest two angles as 220, which is lower than 240. Therefore, 240 [D] is the correct answer.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2013|ab=B|num-b=12|num-a=14}}<br />
<br />
[[Category:Introductory Geometry Problems]]</div>LOTRFan123