Difference between revisions of "1960 AHSME Problems/Problem 38"
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==Problem== | ==Problem== | ||
+ | |||
+ | In this diagram <math>AB</math> and <math>AC</math> are the equal sides of an isosceles <math>\triangle ABC</math>, in which is inscribed equilateral <math>\triangle DEF</math>. | ||
+ | Designate <math>\angle BFD</math> by <math>a</math>, <math>\angle ADE</math> by <math>b</math>, and <math>\angle FEC</math> by <math>c</math>. Then: | ||
+ | |||
+ | <asy> | ||
+ | size(150); | ||
+ | defaultpen(linewidth(0.8)+fontsize(10)); | ||
+ | pair A=(5,12),B=origin,C=(10,0),D=(5/3,4),E=(10-5*.45,12*.45),F=(6,0); | ||
+ | draw(A--B--C--cycle^^D--E--F--cycle); | ||
+ | draw(anglemark(E,D,A,1,45)^^anglemark(F,E,C,1,45)^^anglemark(D,F,B,1,45)); | ||
+ | label("$b$",(D.x+.2,D.y+.25),dir(30)); | ||
+ | label("$c$",(E.x,E.y-.4),S); | ||
+ | label("$a$",(F.x-.4,F.y+.1),dir(150)); | ||
+ | label("$A$",A,N); | ||
+ | label("$B$",B,S); | ||
+ | label("$C$",C,S); | ||
+ | label("$D$",D,dir(150)); | ||
+ | label("$E$",E,dir(60)); | ||
+ | label("$F$",F,S);</asy> | ||
+ | |||
+ | <math>\textbf{(A)}\ b=\frac{a+c}{2}\qquad | ||
+ | \textbf{(B)}\ b=\frac{a-c}{2}\qquad | ||
+ | \textbf{(C)}\ a=\frac{b-c}{2} \qquad | ||
+ | \textbf{(D)}\ a=\frac{b+c}{2}\qquad | ||
+ | \textbf{(E)}\ \text{none of these} </math> | ||
+ | |||
+ | ==Solution== | ||
+ | Since <math>\triangle DEF</math> is an [[equilateral triangle]], all of the angles are <math>60^{\circ}</math>. | ||
+ | The angles in a line add up to <math>180^{\circ}</math>, so | ||
+ | <cmath>\angle FDB = 120 - b</cmath> | ||
+ | <cmath>\angle EFC = 120 - a</cmath> | ||
+ | The angles in a triangle add up to <math>180^{\circ}</math>, so | ||
+ | <cmath>\angle ABC = 60 + b - a</cmath> | ||
+ | <cmath>\angle ACB = 60 - c + a</cmath> | ||
+ | Since <math>\triangle ABC</math> is [[isosceles triangle|isosceles]] and <math>AB = AC</math>, by Base-Angle Theorem, | ||
+ | <cmath>60 + b - a = 60 - c + a</cmath> | ||
+ | <cmath>b + c = 2a</cmath> | ||
+ | <cmath>a = \frac{b+c}{2}</cmath> | ||
+ | The answer is <math>\boxed{\textbf{(D)}}</math>. | ||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/ZdM2ou5Gsuw?t=230 | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
==See Also== | ==See Also== | ||
− | {{AHSME 40p box|year=1960 | | + | {{AHSME 40p box|year=1960|num-b=37|num-a=39}} |
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 21:27, 28 December 2023
Contents
[hide]Problem
In this diagram and are the equal sides of an isosceles , in which is inscribed equilateral . Designate by , by , and by . Then:
Solution
Since is an equilateral triangle, all of the angles are . The angles in a line add up to , so The angles in a triangle add up to , so Since is isosceles and , by Base-Angle Theorem, The answer is .
Video Solution
https://youtu.be/ZdM2ou5Gsuw?t=230
~MathProblemSolvingSkills.com
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 37 |
Followed by Problem 39 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |