Difference between revisions of "2000 AIME I Problems/Problem 14"
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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>. | 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 == | + | == 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(" | <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> | </asy></center> | ||
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<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 | + | == 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>. | 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>. | 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 == |
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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