Difference between revisions of "2011 UNCO Math Contest II Problems/Problem 3"

Revision as of 17:55, 5 May 2019

Problem

The two congruent rectangles shown have dimensions $5$ in. by $25$ in. What is the area of the shaded overlap region? $[asy] filldraw((0,0)--(13,0)--(25,5)--(12,5)--cycle,blue); pair A1=(0,0),B1=(25,0),C1=(25,5),D1=(0,5); draw(A1--B1--C1--D1--cycle,black); pair P,R; P=unit(A1-D1)+D1; R=unit(C1-D1)+D1; draw(P--(P+R-D1)--R,black); pair A2=(0,0),B2=(25/13,-60/13),C2=(25,5),D2=(25-25/13,5+60/13); draw(A2--B2--C2--D2--cycle,black); P=unit(A2-D2)+D2; R=unit(C2-D2)+D2; draw(P--(P+R-D2)--R,black); [/asy]$

Solution

$65$ [asy] filldraw((0,0)--(13,0)--(25,5)--(12,5)--cycle,blue); pair A1=(0,0),B1=(25,0),C1=(25,5),D1=(0,5); draw(A1--B1--C1--D1--cycle,black); pair P,R; P=unit(A1-D1)+D1; R=unit(C1-D1)+D1; draw(P--(P+R-D1)--R,black); pair A2=(0,0),B2=(25/13,-60/13),C2=(25,5),D2=(25-25/13,5+60/13); draw(A2--B2--C2--D2--cycle,black); P=unit(A2-D2)+D2; R=unit(C2-D2)+D2; draw(P--(P+R-D2)--R,black); label("A",(0,0),SW); label("B",(0,5),NW); label("M",(12,5),NW); label("C",(25,5),NE); label("D",(25,0),SE); label("N",(13,0),SE); label("O",(25/13,-60/13),S); label("P",(25-25/13,5+60/13); [/asy] We notice that triangles $AON$ and $NDC$ are congruent, since they share two angles and a side (a right angle, and the opposite angle, and the side 5) This means that if we label $ON$ $x$ then $NC=25-x$ and $AN^2=5^2+x^2$ and so since $AON$ and $NDC$ are congruent, then $AN=NC$ and so $(25-x)^2=5^2+x^2$ solving for $x$ we get 12, meaning that the area of the two triangles $AON$ and $MPC$ (which are also congruent since the $AD$ is parallel to $BC$ ) is 30, so the total area of them is 60, and so taking this away from rectangle $AOCP$ we get the blue, so $125-60=65$ so the answer is 65.

2011 UNCO Math Contest II (Problems • Answer Key • Resources)
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Difference between revisions of "2011 UNCO Math Contest II Problems/Problem 3"

Revision as of 17:55, 5 May 2019

Problem

Solution

See Also

@@ Line 21: / Line 21: @@
 == Solution ==
 <math>65</math>
+[asy] filldraw((0,0)--(13,0)--(25,5)--(12,5)--cycle,blue); pair A1=(0,0),B1=(25,0),C1=(25,5),D1=(0,5);
+draw(A1--B1--C1--D1--cycle,black); pair P,R; P=unit(A1-D1)+D1; R=unit(C1-D1)+D1; draw(P--(P+R-D1)--R,black);
+pair A2=(0,0),B2=(25/13,-60/13),C2=(25,5),D2=(25-25/13,5+60/13); draw(A2--B2--C2--D2--cycle,black);
+P=unit(A2-D2)+D2; R=unit(C2-D2)+D2;
+draw(P--(P+R-D2)--R,black);
+label("A",(0,0),SW);
+label("B",(0,5),NW);
+label("M",(12,5),NW);
+label("C",(25,5),NE);
+label("D",(25,0),SE);
+label("N",(13,0),SE);
+label("O",(25/13,-60/13),S);
+label("P",(25-25/13,5+60/13);
+[/asy]
+We notice that triangles <math>AON</math> and <math>NDC</math> are congruent, since they share two angles and a side (a right angle, and the opposite angle, and the side 5) This means that if we label <math>ON</math> <math>x</math> then <math>NC=25-x</math> and <math>AN^2=5^2+x^2</math> and so since <math>AON</math> and <math>NDC</math> are congruent, then <math>AN=NC</math> and so <math>(25-x)^2=5^2+x^2</math> solving for <math>x</math> we get 12, meaning that the area of the two triangles <math>AON</math> and <math>MPC</math> (which are also congruent since the <math>AD</math> is parallel to <math>BC</math>) is 30, so the total area of them is 60, and so taking this away from rectangle <math>AOCP</math> we get the blue, so <math>125-60=65</math> so the answer is 65.
 == See Also ==