Difference between revisions of "2002 AMC 12B Problems/Problem 24"
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We have | We have | ||
<cmath>[ABCD] = 2002 \le \frac 12 (AC \cdot BD)</cmath> | <cmath>[ABCD] = 2002 \le \frac 12 (AC \cdot BD)</cmath> | ||
− | ( | + | (This is true for any convex quadrilateral: split the quadrilateral along <math>AC</math> and then using the triangle area formula to evaluate <math>[ACB]</math> and <math>[ACD]</math>), with equality only if <math>\overline{AC} \perp \overline{BD}</math>. By the [[triangle inequality]], |
<cmath>\begin{align*}AC &\le PA + PC = 52\\ | <cmath>\begin{align*}AC &\le PA + PC = 52\\ | ||
− | + | BD &\le PB + PD = 77\end{align*}</cmath> | |
with equality if <math>P</math> lies on <math>\overline{AC}</math> and <math>\overline{BD}</math> respectively. Thus | with equality if <math>P</math> lies on <math>\overline{AC}</math> and <math>\overline{BD}</math> respectively. Thus | ||
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<cmath>2002 \le \frac{1}{2} AC \cdot BD \le \frac 12 \cdot 52 \cdot 77 = 2002</cmath> | <cmath>2002 \le \frac{1}{2} AC \cdot BD \le \frac 12 \cdot 52 \cdot 77 = 2002</cmath> | ||
− | Since we have the equality case, <math>\overline{AC} \perp \overline{BD}</math> at point <math>P</math>. | + | Since we have the equality case, <math>\overline{AC} \perp \overline{BD}</math> at point <math>P</math>, as shown below. |
− | + | ||
+ | <center><asy> | ||
+ | size(200); | ||
+ | defaultpen(0.6); | ||
+ | pair A = (0,0), B = (40,0), C = (25.6 * 52 / 24, 19.2 * 52 / 24), D = (40 - (40-25.6)*77/32,19.2*77/32), P = (25.6,19.2), Q = (25.6, 18.5); | ||
+ | pair E=(A+P)/2, F=(B+P)/2, G=(C+P)/2, H=(D+P)/2; | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(A--P--B--P--C--P--D); | ||
+ | label("\(A\)",A,WSW); | ||
+ | label("\(B\)",B,ESE); | ||
+ | label("\(C\)",C,ESE); | ||
+ | label("\(D\)",D,NW); | ||
+ | label("\(P\)",Q,SSW); | ||
+ | label("24",E,WNW); | ||
+ | label("32",F,WSW); | ||
+ | label("28",G,ESE); | ||
+ | label("45",H,ENE); | ||
+ | draw(rightanglemark(C,P,D,50)); | ||
+ | </asy></center> | ||
By the [[Pythagorean Theorem]], | By the [[Pythagorean Theorem]], | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 10:23, 4 July 2013
Problem
A convex quadrilateral with area
contains a point
in its interior such that
. Find the perimeter of
.
Solution
We have
(This is true for any convex quadrilateral: split the quadrilateral along
and then using the triangle area formula to evaluate
and
), with equality only if
. By the triangle inequality,
with equality if lies on
and
respectively. Thus
Since we have the equality case, at point
, as shown below.
![[asy] size(200); defaultpen(0.6); pair A = (0,0), B = (40,0), C = (25.6 * 52 / 24, 19.2 * 52 / 24), D = (40 - (40-25.6)*77/32,19.2*77/32), P = (25.6,19.2), Q = (25.6, 18.5); pair E=(A+P)/2, F=(B+P)/2, G=(C+P)/2, H=(D+P)/2; draw(A--B--C--D--cycle); draw(A--P--B--P--C--P--D); label("\(A\)",A,WSW); label("\(B\)",B,ESE); label("\(C\)",C,ESE); label("\(D\)",D,NW); label("\(P\)",Q,SSW); label("24",E,WNW); label("32",F,WSW); label("28",G,ESE); label("45",H,ENE); draw(rightanglemark(C,P,D,50)); [/asy]](http://latex.artofproblemsolving.com/b/e/0/be0cac31118507011fcd067dec64c89d4f9eee03.png)
By the Pythagorean Theorem,
The perimeter of is
.
See also
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.