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Difference between revisions of "1964 AHSME Problems/Problem 33"
(Added diagram to solution)
 
(2 intermediate revisions by one other user not shown)
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<asy>
 
<asy>
−
draw((0,0)--(6.5,0)--(6.5,4.5)--(0,4.5)--cycle);
+
pair A, B, C, D, P;
−
draw((2.5,1.5)--(0,0));
+
 
−
draw((2.5,1.5)--(0,4.5));
+
A = (0, 0);
−
draw((2.5,1.5)--(6.5,4.5));
+
B = (6.5, 0);
−
draw((2.5,1.5)--(6.5,0),linetype("8 8"));
+
C = (6.5, 4.5);
−
label("$A$",(0,0),dir(-135));
+
D = (0, 4.5);
−
label("$B$",(6.5,0),dir(-45));
+
P = (2.5, 1.5);
−
label("$C$",(6.5,4.5),dir(45));
+
 
−
label("$D$",(0,4.5),dir(135));
+
draw(A--B--C--D--cycle);
−
label("$P$",(2.5,1.5),dir(-90));
+
draw(A--P);
−
label("$3$",(1.25,0.75),dir(120));
+
draw(C--P);
−
label("$4$",(1.25,3),dir(35));
+
draw(D--P);
−
label("$5$",(4.5,3),dir(120));
+
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>
 
</asy>
  
 
==Solution==
 
==Solution==
  
−
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>.
+
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);
 +
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 <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>.
 
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>.
Line 29: Line 79:
 
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>.
 
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>.
  
−
Thus, <math>34 = 16 + x^2</math>, or <math>x = \sqrt{18} = 3\sqrt{2}</math>, which is option <math>\boxed{\textbb{(B)}</math>
+
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 23:46, 14 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 (Problems • Answer Key • Resources)
Preceded by
Problem 32 		Followed by
Problem 34
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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