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

Difference between revisions of "1978 AHSME Problems/Problem 23"

(Created page with "==Problem==")
 
(→‎Problem)
Line 1: Line 1:
 
==Problem==
 
==Problem==
 +
[asy] size(100); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((0,1)--(1,0)); draw((0,0)--(.5,sqrt(3)/2)--(1,0)); label("<math>A</math>",(0,0),SW); label("<math>B</math>",(1,0),SE); label("<math>C</math>",(1,1),NE); label("<math>D</math>",(0,1),NW); label("<math>E</math>",(.5,sqrt(3)/2),E); label("<math>F</math>",intersectionpoint((0,0)--(.5,sqrt(3)/2),(0,1)--(1,0)),2W); //Credit to chezbgone2 for the diagram [/asy]
 +
Vertex <math>E</math> of equilateral <math>\triangle ABE</math> is in the interior of square <math>ABCD</math>, and <math>F</math> is the point of intersection of diagonal <math>BD</math> and line segment <math>AE</math>. If length <math>AB</math> is <math>\sqrt{1+\sqrt{3}}</math> then the area of <math>\triangle ABF</math> is
 +
 +
<math>\textbf{(A) }1\qquad \textbf{(B) }\frac{\sqrt{2}}{2}\qquad \textbf{(C) }\frac{\sqrt{3}}{2}\qquad \textbf{(D) }4-2\sqrt{3}\qquad  \textbf{(E) }\frac{1}{2}+\frac{\sqrt{3}}{4}</math>
 +
 +
==Solution==
 +
No solutions yet!

Revision as of 15:41, 18 June 2021

Problem

[asy] size(100); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((0,1)--(1,0)); draw((0,0)--(.5,sqrt(3)/2)--(1,0)); label("$A$",(0,0),SW); label("$B$",(1,0),SE); label("$C$",(1,1),NE); label("$D$",(0,1),NW); label("$E$",(.5,sqrt(3)/2),E); label("$F$",intersectionpoint((0,0)--(.5,sqrt(3)/2),(0,1)--(1,0)),2W); //Credit to chezbgone2 for the diagram [/asy] Vertex $E$ of equilateral $\triangle ABE$ is in the interior of square $ABCD$, and $F$ is the point of intersection of diagonal $BD$ and line segment $AE$. If length $AB$ is $\sqrt{1+\sqrt{3}}$ then the area of $\triangle ABF$ is

$\textbf{(A) }1\qquad \textbf{(B) }\frac{\sqrt{2}}{2}\qquad \textbf{(C) }\frac{\sqrt{3}}{2}\qquad \textbf{(D) }4-2\sqrt{3}\qquad \textbf{(E) }\frac{1}{2}+\frac{\sqrt{3}}{4}$

Solution

No solutions yet!

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_23&oldid=156315"