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1969 AHSME Problems/Problem 30: Difference between revisions

Timneh (talk | contribs)
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Created page with "== Problem == Let <math>P</math> be a point of hypotenuse <math>AB</math> (or its extension) of isosceles right triangle <math>ABC</math>. Let <math>s=AP^2+PB^2</math>. Then: <..."
 
Rockmanex3 (talk | contribs)
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Solution to Problem 30
 
Line 10: Line 10:


== Solution ==
== Solution ==
<math>\fbox{D}</math>


== See also ==
<asy>
pair A=(0,50),B=(50,0),C=(0,0);
draw(A--B--C--A);
dot(A);
label("$A$",A,NE);
dot(B);
label("$B$",B,NE);
dot(C);
label("$C$",C,SW);
 
pair P=(14,36);
dot(P);
label("$P$",P,NE);
draw((0,36)--P,dotted);
draw((14,0)--P,dotted);
 
label("$a$",(7,36),S);
label("$a$",(0,43),W);
label("$x-a$",(14,18),E);
label("$x-a$",(32,0),S);
</asy>
 
Consider the case where <math>P</math> is on the hypotenuse of <math>AB</math>.  Draw perpendicular lines from <math>P</math> towards the sides.  Using the [[Pythagorean Theorem]],
<cmath>AP^2 = a^2 + a^2</cmath>
<cmath>BP^2 = (x-a)^2 + (x-a)^2</cmath>
<cmath>CP^2 = a^2 + (x-a)^2</cmath>
This means
<cmath>s = 4a^2 - 4ax + 2x^2</cmath>
<cmath>2 \cdot CP^2 = 4a^2 - 4ax + 2x^2</cmath>
Thus, <math>s = 2 \cdot CP^2</math> when <math>P</math> is on the hypotenuse of <math>AB</math>.
 
 
<asy>
pair A=(0,50),B=(50,0),C=(0,0);
draw(A--B--C--A);
dot(A);
label("$A$",A,NE);
dot(B);
label("$B$",B,NE);
dot(C);
label("$C$",C,SW);
 
pair P=(-10,60);
dot(P);
label("$P$",P,NE);
draw(P--A);
draw(P--(-10,50)--A,dotted);
draw(P--(-10,0)--C,dotted);
 
pair D=(-10,50);
dot(D);
label("$D$",D,SW);
 
label("$a$",(-10,55),W);
label("$a$",(-5,50),S);
label("$x$",(0,25),E);
label("$x$",(25,0),S);
 
</asy>
 
Consider the case where <math>P</math> is on the extension of <math>AB</math>.  [[WLOG]], let point <math>A</math> be between point <math>P</math> and point <math>B</math>.  Extend <math>BC</math> and draw perpendicular line from <math>P</math>.  Also, draw point <math>D</math>, where <math>PD \parallel AC</math> and <math>DA \parallel CB</math>.
 
Using the Pythagorean Theorem again,
<cmath>AP^2 = a^2 + a^2</cmath>
<cmath>BP^2 = (a+x)^2 + (a+x)^2</cmath>
<cmath>CP^2 = (a+x)^2 + a^2</cmath>
That means
<cmath>s = 4a^2 + 4ax + 2x^2</cmath>
<cmath>2 \cdot CP^2 = 4a^2 + 4ax + 2x^2</cmath>
Thus, <math>s = 2 \cdot CP^2</math> when <math>P</math> is outside the hypotenuse.
 
In summary, <math>AP^2 + BP^2 = 2 \cdot CP^2</math>, so the answer is <math>\boxed{\textbf{(D)}}</math>.
 
== See Also ==
{{AHSME 35p box|year=1969|num-b=29|num-a=31}}   
{{AHSME 35p box|year=1969|num-b=29|num-a=31}}   


[[Category: Intermediate Geometry Problems]]
[[Category: Introductory Geometry Problems]]
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 22:12, 21 June 2018

Problem

Let $P$ be a point of hypotenuse $AB$ (or its extension) of isosceles right triangle $ABC$. Let $s=AP^2+PB^2$. Then:

$\text{(A) } s<2CP^2 \text{ for a finite number of positions of P}\quad\\ \text{(B) } s<2CP^2 \text{ for an infinite number of positions of P}\quad\\ \text{(C) } s=2CP^2 \text{ only if P is the midpoint or an endpoint of AB}\quad\\ \text{(D) } s=2CP^2 \text{ always}\quad\\ \text{(E) } s>2CP^2 \text{ if P is a trisection point of AB}$

Solution

[asy] pair A=(0,50),B=(50,0),C=(0,0); draw(A--B--C--A); dot(A); label("$A$",A,NE); dot(B); label("$B$",B,NE); dot(C); label("$C$",C,SW); pair P=(14,36); dot(P); label("$P$",P,NE); draw((0,36)--P,dotted); draw((14,0)--P,dotted); label("$a$",(7,36),S); label("$a$",(0,43),W); label("$x-a$",(14,18),E); label("$x-a$",(32,0),S); [/asy]

Consider the case where $P$ is on the hypotenuse of $AB$. Draw perpendicular lines from $P$ towards the sides. Using the Pythagorean Theorem, \[AP^2 = a^2 + a^2\] \[BP^2 = (x-a)^2 + (x-a)^2\] \[CP^2 = a^2 + (x-a)^2\] This means \[s = 4a^2 - 4ax + 2x^2\] \[2 \cdot CP^2 = 4a^2 - 4ax + 2x^2\] Thus, $s = 2 \cdot CP^2$ when $P$ is on the hypotenuse of $AB$.


[asy] pair A=(0,50),B=(50,0),C=(0,0); draw(A--B--C--A); dot(A); label("$A$",A,NE); dot(B); label("$B$",B,NE); dot(C); label("$C$",C,SW); pair P=(-10,60); dot(P); label("$P$",P,NE); draw(P--A); draw(P--(-10,50)--A,dotted); draw(P--(-10,0)--C,dotted); pair D=(-10,50); dot(D); label("$D$",D,SW); label("$a$",(-10,55),W); label("$a$",(-5,50),S); label("$x$",(0,25),E); label("$x$",(25,0),S); [/asy]

Consider the case where $P$ is on the extension of $AB$. WLOG, let point $A$ be between point $P$ and point $B$. Extend $BC$ and draw perpendicular line from $P$. Also, draw point $D$, where $PD \parallel AC$ and $DA \parallel CB$.

Using the Pythagorean Theorem again, \[AP^2 = a^2 + a^2\] \[BP^2 = (a+x)^2 + (a+x)^2\] \[CP^2 = (a+x)^2 + a^2\] That means \[s = 4a^2 + 4ax + 2x^2\] \[2 \cdot CP^2 = 4a^2 + 4ax + 2x^2\] Thus, $s = 2 \cdot CP^2$ when $P$ is outside the hypotenuse.

In summary, $AP^2 + BP^2 = 2 \cdot CP^2$, so the answer is $\boxed{\textbf{(D)}}$.

See Also

1969 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 31
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
All AHSME Problems and Solutions

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