Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

1964 AHSME Problems/Problem 29: Difference between revisions

← Older editNewer edit →
Talkinaway (talk | contribs)
680 edits
No edit summary
Talkinaway (talk | contribs)
680 edits
Line 29: Line 29:
We examine <math>\triangle RDF</math> and <math>\triangle SFR</math>.  We are given <math>\angle RDF \cong \angle SFR</math>. Also note that <math>\frac{SF}{RD} = \frac{7.5}{6} = 1.25</math> and <math>\frac{FR}{DF} = \frac{5}{4} = 1.25</math>, so <math>\frac{SF}{RD} = \frac{FR}{DF}</math>.
We examine <math>\triangle RDF</math> and <math>\triangle SFR</math>.  We are given <math>\angle RDF \cong \angle SFR</math>. Also note that <math>\frac{SF}{RD} = \frac{7.5}{6} = 1.25</math> and <math>\frac{FR}{DF} = \frac{5}{4} = 1.25</math>, so <math>\frac{SF}{RD} = \frac{FR}{DF}</math>.


If two triangles have two pairs of sides that are proportional, and the included angle is congruent, then the two triangles are similar by SAS congruence.  Therefore, the third pair of sides must also be in the same proportion, so  
If two triangles have two pairs of sides that are proportional, and the included angles are congruent, then the two triangles are similar by SAS congruence.  Therefore, the third pair of sides must also be in the same proportion, so  


<math>\frac{SF}{RD} = \frac{FR}{DF} = \frac{RS}{FR} = 1.25</math>  
<math>\frac{SF}{RD} = \frac{FR}{DF} = \frac{RS}{FR} = 1.25</math>  

Revision as of 22:35, 24 July 2019

Problem

In this figure $\angle RFS=\angle FDR$, $FD=4$ inches, $DR=6$ inches, $FR=5$ inches, $FS=7\dfrac{1}{2}$ inches. The length of $RS$, in inches, is:

$\textbf{(A) }\text{undetermined}\qquad\textbf{(B) }4\qquad\textbf{(C) }5\dfrac{1}{2}\qquad\textbf{(D) }6\qquad\textbf{(E) }6\dfrac{1}{2}\qquad$

[asy] import olympiad; draw((0,0)--7.5*dir(0)); draw((0,0)--5*dir(55)); draw((0,0)--4*dir(140)); draw(4*dir(140)--5*dir(55)--7.5*dir(0)); markscalefactor=0.1; draw(anglemark((0,0),4*dir(140),5*dir(55))); draw(anglemark(7.5*dir(0),(0,0),5*dir(55))); label("$F$",(0,0),dir(240)); label("$D$",4*dir(140),dir(180)); label("$R$",5*dir(55),dir(80)); label("$S$",7.5*dir(0),dir(-25)); label("$4$",2*dir(140),dir(230)); label("$5$",2.5*dir(55),dir(155)); label("$6$",(4*dir(140)+5*dir(55))/2,dir(100)); label("$7\frac{1}{2}$",3.75,dir(-90)); [/asy]

Solution

We examine $\triangle RDF$ and $\triangle SFR$. We are given $\angle RDF \cong \angle SFR$. Also note that $\frac{SF}{RD} = \frac{7.5}{6} = 1.25$ and $\frac{FR}{DF} = \frac{5}{4} = 1.25$, so $\frac{SF}{RD} = \frac{FR}{DF}$.

If two triangles have two pairs of sides that are proportional, and the included angles are congruent, then the two triangles are similar by SAS congruence. Therefore, the third pair of sides must also be in the same proportion, so

$\frac{SF}{RD} = \frac{FR}{DF} = \frac{RS}{FR} = 1.25$

$\frac{RS}{FR} = 1.25$

$\frac{RS}{5} = 1.25$

$RS = 6.25$, which is answer $\boxed{\textbf{(E) }}$

See Also

1964 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_29&oldid=107908"