Difference between revisions of "1964 AHSME Problems/Problem 33"

(Added diagram to solution)
 
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<asy>
 
<asy>
draw((0,0)--(6.5,0)--(6.5,4.5)--(0,4.5)--cycle);
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pair A, B, C, D, P;
draw((2.5,1.5)--(0,0));
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draw((2.5,1.5)--(0,4.5));
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A = (0, 0);
draw((2.5,1.5)--(6.5,4.5));
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B = (6.5, 0);
draw((2.5,1.5)--(6.5,0),linetype("8 8"));
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C = (6.5, 4.5);
label("$A$",(0,0),dir(-135));
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D = (0, 4.5);
label("$B$",(6.5,0),dir(-45));
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P = (2.5, 1.5);
label("$C$",(6.5,4.5),dir(45));
+
 
label("$D$",(0,4.5),dir(135));
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draw(A--B--C--D--cycle);
label("$P$",(2.5,1.5),dir(-90));
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draw(A--P);
label("$3$",(1.25,0.75),dir(120));
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draw(C--P);
label("$4$",(1.25,3),dir(35));
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draw(D--P);
label("$5$",(4.5,3),dir(120));
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draw(B--P, dashed);
 +
 
 +
label("$A$", A, SW);
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label("$B$", B, SE);
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label("$C$", C, NE);
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label("$D$", D, NW);
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label("$P$", P, S);
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label("$3$", midpoint(A--P), NW);
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label("$4$", midpoint(D--P), NE);
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label("$5$", midpoint(C--P), NW);
 
</asy>
 
</asy>
 +
 +
==Solution==
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 +
From point <math>P</math>, create perpendiculars to all four sides, labeling them <math>a, b, c, d</math> starting from going north and continuing clockwise.  Label the length <math>PB</math> as <math>x</math>:
 +
 +
<asy>
 +
unitsize(1cm);
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pair A, B, C, D, P, ABfoot, BCfoot, CDfoot, DAfoot;
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 +
A = (0, 0);
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B = (6.5, 0);
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C = (6.5, 4.5);
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D = (0, 4.5);
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P = (2.5, 1.5);
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ABfoot = (2.5, 0);
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BCfoot = (6.5, 1.5);
 +
CDfoot = (2.5, 4.5);
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DAfoot = (0, 1.5);
 +
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draw(A--B--C--D--cycle);
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draw(A--P);
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draw(C--P);
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draw(D--P);
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draw(B--P, dashed);
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draw(ABfoot--CDfoot);
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draw(DAfoot--BCfoot);
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draw(rightanglemark(P, CDfoot, D));
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draw(rightanglemark(P, BCfoot, C));
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draw(rightanglemark(P, ABfoot, B));
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draw(rightanglemark(P, DAfoot, A));
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label("$A$", A, SW);
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label("$B$", B, SE);
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label("$C$", C, NE);
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label("$D$", D, NW);
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label("$P$", P, 3*dir(240));
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label("$3$", midpoint(A--P), NW);
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label("$4$", midpoint(D--P), NE);
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label("$5$", midpoint(C--P), NW);
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label("$x$", midpoint(B--P), SW);
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label("$a$", midpoint(P--CDfoot), E);
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label("$b$", midpoint(P--BCfoot), N);
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label("$c$", midpoint(P--ABfoot), E);
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label("$d$", midpoint(P--DAfoot), N);
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</asy>
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 +
We have <math>a^2 + b^2 = 5^2</math> and <math>c^2 + d^2 = 3^2</math>, leading to <math>a^2 + b^2 + c^2 + d^2 = 34</math>.
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We also have <math>a^2 + d^2 = 4^2</math> and <math>b^2 + c^2 = x^2</math>, leading to <math>a^2 + b^2 + c^2 + d^2 = 16 + x^2</math>.
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Thus, <math>34 = 16 + x^2</math>, or <math>x = \sqrt{18} = 3\sqrt{2}</math>, which is option <math>\boxed{\textbf{(B)}}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 00:46, 15 June 2022

Problem

$P$ is a point interior to rectangle $ABCD$ and such that $PA=3$ inches, $PD=4$ inches, and $PC=5$ inches. Then $PB$, in inches, equals:

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

[asy] pair A, B, C, D, P;  A = (0, 0); B = (6.5, 0); C = (6.5, 4.5); D = (0, 4.5); P = (2.5, 1.5);  draw(A--B--C--D--cycle); draw(A--P); draw(C--P); draw(D--P); draw(B--P, dashed);  label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$P$", P, S); label("$3$", midpoint(A--P), NW); label("$4$", midpoint(D--P), NE); label("$5$", midpoint(C--P), NW); [/asy]

Solution

From point $P$, create perpendiculars to all four sides, labeling them $a, b, c, d$ starting from going north and continuing clockwise. Label the length $PB$ as $x$:

[asy] unitsize(1cm); pair A, B, C, D, P, ABfoot, BCfoot, CDfoot, DAfoot;  A = (0, 0); B = (6.5, 0); C = (6.5, 4.5); D = (0, 4.5); P = (2.5, 1.5); ABfoot = (2.5, 0); BCfoot = (6.5, 1.5); CDfoot = (2.5, 4.5); DAfoot = (0, 1.5);  draw(A--B--C--D--cycle); draw(A--P); draw(C--P); draw(D--P); draw(B--P, dashed); draw(ABfoot--CDfoot); draw(DAfoot--BCfoot); draw(rightanglemark(P, CDfoot, D)); draw(rightanglemark(P, BCfoot, C)); draw(rightanglemark(P, ABfoot, B)); draw(rightanglemark(P, DAfoot, A));  label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$P$", P, 3*dir(240)); label("$3$", midpoint(A--P), NW); label("$4$", midpoint(D--P), NE); label("$5$", midpoint(C--P), NW); label("$x$", midpoint(B--P), SW); label("$a$", midpoint(P--CDfoot), E); label("$b$", midpoint(P--BCfoot), N); label("$c$", midpoint(P--ABfoot), E); label("$d$", midpoint(P--DAfoot), N); [/asy]

We have $a^2 + b^2 = 5^2$ and $c^2 + d^2 = 3^2$, leading to $a^2 + b^2 + c^2 + d^2 = 34$.

We also have $a^2 + d^2 = 4^2$ and $b^2 + c^2 = x^2$, leading to $a^2 + b^2 + c^2 + d^2 = 16 + x^2$.

Thus, $34 = 16 + x^2$, or $x = \sqrt{18} = 3\sqrt{2}$, which is option $\boxed{\textbf{(B)}}$

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 32
Followed by
Problem 34
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