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Difference between revisions of "1963 AHSME Problems/Problem 18"

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//Credit to MSTang for the asymptote</asy>
 
//Credit to MSTang for the asymptote</asy>
  
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If we let <math>D</math> be on <math>FE</math> so that <math>FD = DB</math>, then by SAS Congruency, <math>\triangle DMB \cong \triangle DMC</math>, so <math>FD = DB = DC</math>.  Since three points define a circle and point <math>D</math> is equidistant from three points, <math>D</math> is the center, so <math>FE</math> is a diameter.  Therefore, <math>\angle EFA</math> is a right angle, and by AA Similarity, we can confirm that <math>\triangle EFA \sim \triangle EUM</math>.  The answer is <math>\boxed{\textbf{(A)}}</math>.
+
If we let <math>D</math> be on <math>FE</math> so that <math>FD = DB</math>, then by SAS Congruency, <math>\triangle DMB \cong \triangle DMC</math>, so <math>FD = DB = DC</math>.  Since three points define a circle and point <math>D</math> is equidistant from three points, <math>D</math> is the center, so <math>FE</math> is a diameter.  Therefore, <math>\angle EAF</math> is a right angle, and by AA Similarity, we can confirm that <math>\triangle EFA \sim \triangle EUM</math>.  The answer is <math>\boxed{\textbf{(A)}}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 20:18, 31 May 2020

Problem

Chord $EF$ is the perpendicular bisector of chord $BC$, intersecting it in $M$. Between $B$ and $M$ point $U$ is taken, and $EU$ extended meets the circle in $A$. Then, for any selection of $U$, as described, $\triangle EUM$ is similar to:

[asy] pair B = (-0.866, -0.5); pair C = (0.866, -0.5); pair E = (0, -1); pair F = (0, 1); pair M = midpoint(B--C); pair A = (-0.99, -0.141); pair U = intersectionpoints(A--E, B--C)[0]; draw(B--C); draw(F--E--A); draw(unitcircle); label("$B$", B, SW); label("$C$", C, SE); label("$A$", A, W); label("$E$", E, S); label("$U$", U, NE); label("$M$", M, NE); label("$F$", F, N); //Credit to MSTang for the asymptote[/asy]

$\textbf{(A)}\ \triangle EFA \qquad \textbf{(B)}\ \triangle EFC \qquad \textbf{(C)}\ \triangle ABM \qquad \textbf{(D)}\ \triangle ABU \qquad \textbf{(E)}\ \triangle FMC$

Solution

Note that $\triangle EUM$ is a right triangle with one of the angles being $\angle FEA$. This leads to prediction that $\triangle FEA$ is the similar triangle as it shares an angle, and to prove this, we need to show that $FE$ is a diameter.

[asy] pair B = (-0.866, -0.5); pair C = (0.866, -0.5); pair E = (0, -1); pair F = (0, 1); pair M = midpoint(B--C); pair A = (-0.99, -0.141); pair U = intersectionpoints(A--E, B--C)[0]; draw(B--C); draw(F--E--A); draw(unitcircle); label("$B$", B, SW); label("$C$", C, SE); label("$A$", A, W); label("$E$", E, S); label("$U$", U, NE); label("$M$", M, NE); label("$F$", F, N); draw("$D$",(0,0),NE); draw(B--(0,0)--C,dotted); draw(F--A); dot(A); dot(B); dot(C); dot((0,0)); dot(E); dot(F); dot(U); dot(M); //Credit to MSTang for the asymptote[/asy]

If we let $D$ be on $FE$ so that $FD = DB$, then by SAS Congruency, $\triangle DMB \cong \triangle DMC$, so $FD = DB = DC$. Since three points define a circle and point $D$ is equidistant from three points, $D$ is the center, so $FE$ is a diameter. Therefore, $\angle EAF$ is a right angle, and by AA Similarity, we can confirm that $\triangle EFA \sim \triangle EUM$. The answer is $\boxed{\textbf{(A)}}$.

See Also

1963 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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

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