Difference between revisions of "2000 AIME I Problems/Problem 14"
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== Problem == | == Problem == | ||
− | In triangle <math>ABC,</math> it is given that angles <math>B</math> and <math>C</math> are congruent. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AC}</math> and <math>\overline{AB},</math> respectively, so that <math>AP = PQ = QB = BC.</math> Angle <math>ACB</math> is <math>r</math> times as large as angle <math>APQ,</math> where <math>r</math> is a positive real number. Find the | + | In triangle <math>ABC,</math> it is given that angles <math>B</math> and <math>C</math> are [[congruent (geometry)|congruent]]. Points <math>P</math> and <math>Q</math> lie on <math>\overline{AC}</math> and <math>\overline{AB},</math> respectively, so that <math>AP = PQ = QB = BC.</math> Angle <math>ACB</math> is <math>r</math> times as large as angle <math>APQ,</math> where <math>r</math> is a positive real number. Find <math>\lfloor 1000r \rfloor</math>. |
+ | |||
+ | __TOC__ | ||
+ | == Official Solution (MAA)== | ||
+ | |||
+ | <center><asy>defaultpen(fontsize(10)); size(200); pen p=fontsize(8); | ||
+ | pair A,B,C,P,Q; | ||
+ | B=MP("B",origin,down+left); C=MP("C",20*right,right+down); A=MP("A",extension(B,dir(80),C,C+dir(100)),up); Q=MP("Q",20*dir(80),left); P=MP("P",Q+(20*dir(60)),right); | ||
+ | draw(A--B--C--A, black+1);draw(B--P--Q); MP("x",B,20*dir(75),p); MP("x",P,17*dir(245),p); MP("2x",Q,15*dir(70),p); MP("2x",A,15*dir(-90),p); MP("2y",P,2*left,p); MP("3x",P,10*dir(-95),p); MP("x+y",C,5*dir(135),p); MP("y",B,5*dir(40),p); | ||
+ | </asy></center> | ||
+ | Let <math>\angle QPB=x^\circ</math>. Because <math>\angle AQP</math> is exterior to isosceles triangle <math>PQB</math> its measure is <math>2x</math> and <math>\angle PAQ</math> has the same measure. Because <math>\angle BPC</math> is exterior to <math>\triangle BPA</math> its measure is <math>3x</math>. Let <math>\angle PBC = y^\circ</math>. It follows that <math>\angle ACB = x+y</math> and that <math>4x+2y=180^\circ</math>. Two of the angles of triangle <math>APQ</math> have measure <math>2x</math>, and thus the measure of <math>\angle APQ</math> is <math>2y</math>. It follows that <math>AQ=2\cdot AP\cdot \sin y</math>. Because <math>AB=AC</math> and <math>AP=QB</math>, it also follows that <math>AQ=PC</math>. Now apply the Law of Sines to triangle <math>PBC</math> to find <cmath>\frac{\sin 3x}{BC}=\frac{\sin y}{PC}=\frac{\sin y}{2\cdot AP\cdot \sin y}= \frac{1}{2\cdot BC}</cmath>because <math>AP=BC</math>. Hence <math>\sin 3x = \tfrac 12</math>. Since <math>4x<180</math>, this implies that <math>3x=30</math>, i.e. <math>x=10</math>. Thus <math>y=70</math> and <cmath>r=\frac{10+70}{2\cdot 70}=\frac 47,</cmath>which implies that <math>1000r = 571 + \tfrac 37</math>. So the answer is <math>\boxed{571}</math>. | ||
+ | |||
+ | == Solution 1 == | ||
+ | <center><asy>defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60); draw(A--B--C--A);draw(P--Q);draw(A--R--B);draw(P--R);D(R--C,dashed); label(" | ||
+ | </asy></center> | ||
+ | Let point <math>R</math> be in <math>\triangle ABC</math> such that <math>QB = BR = RP</math>. Then <math>PQBR</math> is a [[rhombus]], so <math>AB \parallel PR</math> and <math>APRB</math> is an [[isosceles trapezoid]]. Since <math>\overline{PB}</math> bisects <math>\angle QBR</math>, it follows by symmetry in trapezoid <math>APRB</math> that <math>\overline{RA}</math> bisects <math>\angle BAC</math>. Thus <math>R</math> lies on the perpendicular bisector of <math>\overline{BC}</math>, and <math>BC = BR = RC</math>. Hence <math>\triangle BCR</math> is an [[equilateral triangle]]. | ||
+ | |||
+ | Now <math>\angle ABR = \angle BAC = \angle ACR</math>, and the sum of the angles in <math>\triangle ABC</math> is <math>\angle ABR + 60^{\circ} + \angle BAC + \angle ACR + 60^{\circ} = 3\angle BAC + 120^{\circ} = 180^{\circ} \Longrightarrow \angle BAC = 20^{\circ}</math>. Then <math>\angle APQ = 140^{\circ}</math> and <math>\angle ACB = 80^{\circ}</math>, so the answer is <math>\left\lfloor 1000 \cdot \frac{80}{140} \right\rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = \boxed{571}</math>. | ||
+ | |||
+ | == Solution 2 (Law of sines) == | ||
+ | |||
+ | Let <math>AP=PQ=QB=BC=x</math> and <math>A</math> be the measure of <math>\angle BAC</math>. Since <math>\triangle APQ</math> and <math>\triangle ABC</math> are isoceles, <math>\angle APQ = 180-2A</math> and <math>\angle ACB = 90-\frac{A}{2}</math>. | ||
+ | Because <math>\triangle APQ</math> and <math>\triangle ABC</math> both have a side length <math>x</math> opposite <math>\angle BAC</math>, by the law of sines: | ||
+ | |||
+ | <math>\frac{x}{\sin A}=\frac{AQ}{\sin(180-2A)}=\frac{AQ+x}{\sin(90-\frac{A}{2})}</math> | ||
+ | |||
+ | Simplifying, this becomes | ||
+ | |||
+ | <math>\frac{x}{\sin A}=\frac{AQ}{\sin 2A}=\frac{AQ+x}{\cos \frac{A}{2}}</math> | ||
+ | |||
+ | From the first two fractions, | ||
+ | |||
+ | <math>AQ\cdot \sin A = x \cdot \sin 2A = x \cdot (2\sin A \cos A) \Longrightarrow AQ=x\cdot 2\cos A</math> | ||
+ | |||
+ | Substituting, we have from the first and third fractions, | ||
+ | |||
+ | <math>\frac{x}{\sin A}=\frac{x\cdot 2\cos A + x}{\cos \frac{A}{2}} \Longrightarrow 2\cos A\sin A + \sin A=\sin 2A + \sin A = \cos \frac{A}{2}</math> | ||
+ | |||
+ | By sum-to-product, | ||
+ | |||
+ | <math>\sin 2A + \sin A = 2\sin \frac{3A}{2} \cos \frac{A}{2}</math> | ||
+ | |||
+ | Thus, | ||
+ | |||
+ | <math>2\sin \frac{3A}{2} \cos \frac{A}{2} = \cos \frac{A}{2} \Longrightarrow \sin \frac{3A}{2} = \frac{1}{2}</math> | ||
+ | |||
+ | Because <math>BC=QB<AB</math>, <math>\angle A</math> is acute, so <math>\frac{3A}{2}=30 \Longrightarrow A=20</math> | ||
+ | |||
+ | <math>\angle ACB = \frac{180-20}{2}=80</math>, <math>\angle APQ = 180-2\cdot 20 = 140 \Longrightarrow r=\frac{4}{7}</math> | ||
+ | |||
+ | <math>1000r=\frac{4000}{7}=\boxed{571}.\overline{428571}</math> | ||
+ | |||
+ | ~bad_at_mathcounts | ||
+ | |||
+ | |||
+ | == Solution 3 == | ||
+ | <center><asy>defaultpen(fontsize(8)); size(200); pair A=20*dir(80)+20*dir(60)+20*dir(100), B=(0,0), C=20*dir(0), P=20*dir(80)+20*dir(60), Q=20*dir(80), R=20*dir(60), S; S=intersectionpoint(Q--C,P--B); draw(A--B--C--A);draw(B--P--Q--C--R--Q);draw(A--R--B);draw(P--R--S); label(" | ||
+ | </asy></center> | ||
+ | Again, construct <math>R</math> as above. | ||
− | |||
− | |||
− | |||
Let <math>\angle BAC = \angle QBR = \angle QPR = 2x</math> and <math>\angle ABC = \angle ACB = y</math>, which means <math>x + y = 90</math>. | Let <math>\angle BAC = \angle QBR = \angle QPR = 2x</math> and <math>\angle ABC = \angle ACB = y</math>, which means <math>x + y = 90</math>. | ||
<math>\triangle QBC</math> is isosceles with <math>QB = BC</math>, so <math>\angle BCQ = 90 - \frac {y}{2}</math>. | <math>\triangle QBC</math> is isosceles with <math>QB = BC</math>, so <math>\angle BCQ = 90 - \frac {y}{2}</math>. | ||
− | Let <math>S</math> be the intersection of <math>QC</math>and <math> | + | Let <math>S</math> be the intersection of <math>QC</math> and <math>BP</math>. Since <math>\angle BCQ = \angle BQC = \angle BRS</math>, <math>BCRS</math> is [[cyclic quadrilateral|cyclic]], which means <math>\angle RBS = \angle RCS = x</math>. |
− | Since <math>APRB</math> is an isosceles trapezoid, <math>BP = AR</math>, but since <math>AR</math> bisects <math>\angle BAC</math>, <math>\angle ABR = \angle ACR = 2x</math>. | + | Since <math>APRB</math> is an isosceles trapezoid, <math>BP = AR</math>, but since <math>AR</math> bisects <math>\angle BAC</math>, <math>\angle ABR = \angle ACR = 2x</math>. |
+ | |||
Therefore we have that <math>\angle ACB = \angle ACR + \angle RCS + \angle QCB = 2x + x + 90 - \frac {y}{2} = y</math>. | Therefore we have that <math>\angle ACB = \angle ACR + \angle RCS + \angle QCB = 2x + x + 90 - \frac {y}{2} = y</math>. | ||
We solve the simultaneous equations <math>x + y = 90</math> and <math>2x + x + 90 - \frac {y}{2} = y</math> to get <math>x = 10</math> and <math>y = 80</math>. | We solve the simultaneous equations <math>x + y = 90</math> and <math>2x + x + 90 - \frac {y}{2} = y</math> to get <math>x = 10</math> and <math>y = 80</math>. | ||
<math>\angle APQ = 180 - 4x = 140</math>, <math>\angle ACB = 80</math>, so <math>r = \frac {80}{140} = \frac {4}{7}</math>. | <math>\angle APQ = 180 - 4x = 140</math>, <math>\angle ACB = 80</math>, so <math>r = \frac {80}{140} = \frac {4}{7}</math>. | ||
− | <math>\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}</math> | + | <math>\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}</math>. |
+ | |||
+ | == Solution 4 (Trig identities)== | ||
+ | |||
+ | Let <math>\angle BAC= 2\theta</math> and <math>AP=PQ=QB=BC=x</math>. <math>\triangle APQ</math> is isosceles, so <math>AQ=2x\cos 2\theta =2x(1-2\sin^2\theta)</math> and <math>AB= AQ+x=x\left(3-4\sin^2\theta\right)</math>. <math>\triangle{ABC}</math> is isosceles too, so <math>x=BC=2AB\sin\theta</math>. Using the expression for <math>AB</math>, we get <cmath>1=2\left(3\sin\theta-4\sin^3\theta\right)=2\sin3\theta</cmath>by the triple angle formula! Thus <math>\theta=10^\circ</math> and <math>\angle A = 2\theta=20^\circ</math>. | ||
+ | It follows now that <math>\angle APQ=140^\circ</math>, <math>\angle ACB=80^\circ</math>, giving <math>r=\tfrac{4}{7}</math>, which implies that <math>1000r = 571 + \tfrac 37</math>. So the answer is <math>\boxed{571}</math>. | ||
+ | |||
== See also == | == See also == | ||
− | {{AIME box|year=2000|n=I|num-b=13|num-a=15}} | + | {{AIME box|year=2000|n=I|num-b=13|num-a=15|t=721509}} |
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 02:04, 27 February 2022
Problem
In triangle it is given that angles and are congruent. Points and lie on and respectively, so that Angle is times as large as angle where is a positive real number. Find .
Contents
[hide]Official Solution (MAA)
Let . Because is exterior to isosceles triangle its measure is and has the same measure. Because is exterior to its measure is . Let . It follows that and that . Two of the angles of triangle have measure , and thus the measure of is . It follows that . Because and , it also follows that . Now apply the Law of Sines to triangle to find because . Hence . Since , this implies that , i.e. . Thus and which implies that . So the answer is .
Solution 1
Let point be in such that . Then is a rhombus, so and is an isosceles trapezoid. Since bisects , it follows by symmetry in trapezoid that bisects . Thus lies on the perpendicular bisector of , and . Hence is an equilateral triangle.
Now , and the sum of the angles in is . Then and , so the answer is .
Solution 2 (Law of sines)
Let and be the measure of . Since and are isoceles, and . Because and both have a side length opposite , by the law of sines:
Simplifying, this becomes
From the first two fractions,
Substituting, we have from the first and third fractions,
By sum-to-product,
Thus,
Because , is acute, so
,
~bad_at_mathcounts
Solution 3
Again, construct as above.
Let and , which means . is isosceles with , so . Let be the intersection of and . Since , is cyclic, which means . Since is an isosceles trapezoid, , but since bisects , .
Therefore we have that . We solve the simultaneous equations and to get and . , , so . .
Solution 4 (Trig identities)
Let and . is isosceles, so and . is isosceles too, so . Using the expression for , we get by the triple angle formula! Thus and . It follows now that , , giving , which implies that . So the answer is .
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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