Difference between revisions of "1958 AHSME Problems/Problem 47"

Latest revision as of 14:07, 28 August 2022

Problem

$ABCD$ is a rectangle (see the accompanying diagram) with $P$ any point on $\overline{AB}$ . $\overline{PS} \perp \overline{BD}$ and $\overline{PR} \perp \overline{AC}$ . $\overline{AF} \perp \overline{BD}$ and $\overline{PQ} \perp \overline{AF}$ . Then $PR + PS$ is equal to:

$[asy] draw((-2,-1)--(-2,1)--(2,1)--(2,-1)--cycle,dot); draw((-2,-1)--(2,1)--(2,-1)--(-2,1),dot); draw((-2,1)--(-6/5,-3/5),black+linewidth(.75)); draw((6/5,3/5)--(1,1)--(-3/2+1/10,-2/10),black+linewidth(.75)); draw((1,1)--(1-3/5,1-6/5),black+linewidth(.75)); MP("A",(-2,1),NW);MP("B",(2,1),NE);MP("C",(2,-1),SE);MP("D",(-2,-1),SW); MP("Q",(-3/2+1/10,-2/10),W);MP("T",(-2/5,1/5),N);MP("P",(1,1),N); MP("F",(-6/5,-3/5),SE);MP("E",(0,0),S);MP("S",(6/5,3/5),S);MP("R",(1-3/5,1-6/5),S); [/asy]$

$\textbf{(A)}\ PQ\qquad \textbf{(B)}\ AE\qquad \textbf{(C)}\ PT + AT\qquad \textbf{(D)}\ AF\qquad \textbf{(E)}\ EF$

\usepackage{amssymb}

Solution

Since $\overline{\rm PQ}$ and $\overline{\rm BD}$ are both perpendicular to $\overline{\rm AF}$ , $\overline{\rm PQ} || \overline{\rm BD}$ . Thus, $\angle APQ = \angle ABD$ .

Also, $\angle ABD$ = $\angle CAB$ because $ABCD$ is a rectangle. Thus, $\angle APQ = \angle ABD = \angle CAB$ .

Since $\angle APQ = \angle CAB$ , $\triangle APT$ is isosceles with $PT = AT$ .

$\angle ATQ$ and $\angle PTR$ are vertical angles and are congruent. Thus, by HL congruence, $\triangle ATQ \cong \triangle PTR$ , so $AQ = PR$ .

We also know that $PSFQ$ is a rectangle, since $\overline{\rm PS} \perp \overline{\rm BD}$ , $\overline{\rm BF} \perp \overline{\rm AF}$ , and $\overline{\rm PQ} \perp \overline{\rm AF}$ .

Since $PSFQ$ is a rectangle, $QF = PS$ . We also found earlier that $AQ = PR$ . Thus, $PR + PS = AQ + QF = AF$ .

Our answer is $PR + PS = \fbox{D) AF}$

~ lovesummer

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 46	Followed by Problem 48
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 18: / Line 18: @@
 \textbf{(D)}\ AF\qquad
 \textbf{(E)}\ EF</math>
+\usepackage{amssymb}
 == Solution ==
-<math>\fbox{}</math>
+Since <math>\overline{\rm PQ}</math> and <math>\overline{\rm BD}</math> are both perpendicular to <math>\overline{\rm AF}</math>, <math>\overline{\rm PQ} || \overline{\rm BD}</math>. Thus, <math>\angle APQ = \angle ABD</math>.
+Also, <math>\angle ABD</math> = <math>\angle CAB</math> because <math>ABCD</math> is a rectangle. Thus, <math>\angle APQ = \angle ABD = \angle CAB</math>.
+Since <math>\angle APQ = \angle CAB</math>, <math>\triangle APT</math> is isosceles with <math>PT = AT</math>.
+<math>\angle ATQ</math> and <math>\angle PTR</math> are vertical angles and are congruent. Thus, by HL congruence, <math>\triangle ATQ \cong \triangle PTR</math>, so <math>AQ = PR</math>.
+We also know that <math>PSFQ</math> is a rectangle, since <math>\overline{\rm PS} \perp \overline{\rm BD}</math>, <math>\overline{\rm BF} \perp \overline{\rm AF}</math>, and <math>\overline{\rm PQ} \perp \overline{\rm AF}</math>.
+Since <math>PSFQ</math> is a rectangle, <math>QF = PS</math>. We also found earlier that <math>AQ = PR</math>. Thus, <math>PR + PS = AQ + QF = AF</math>.
+Our answer is <math>PR + PS = \fbox{D) AF}</math>
+~ lovesummer
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Difference between revisions of "1958 AHSME Problems/Problem 47"

Latest revision as of 14:07, 28 August 2022

Problem

Solution

See Also