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Difference between revisions of "2019 AIME II Problems/Problem 1"

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 +
==Problem==
 +
Two different points, <math>C</math> and <math>D</math>, lie on the same side of line <math>AB</math> so that <math>\triangle ABC</math> and <math>\triangle BAD</math> are congruent with <math>AB = 9</math>, <math>BC=AD=10</math>, and <math>CA=DB=17</math>. The intersection of these two triangular regions has area <math>\tfrac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 +
==Solution==
 +
<asy>
 +
unitsize(10);
 +
pair A = (0,0);
 +
pair B = (9,0);
 +
pair C = (15,8);
 +
pair D = (-6,8);
 +
pair E = (-6,0);
 +
draw(A--B--C--cycle);
 +
draw(B--D--A);
 +
label("$A$",A,dir(-120));
 +
label("$B$",B,dir(-60));
 +
label("$C$",C,dir(60));
 +
label("$D$",D,dir(120));
 +
label("$E$",E,dir(-135));
 +
label("$9$",(A+B)/2,dir(-90));
 +
label("$10$",(D+A)/2,dir(-150));
 +
label("$10$",(C+B)/2,dir(-30));
 +
label("$17$",(D+B)/2,dir(60));
 +
label("$17$",(A+C)/2,dir(120));
 +
 +
draw(D--E--A,dotted);
 +
label("$8$",(D+E)/2,dir(180));
 +
label("$6$",(A+E)/2,dir(-90));
 +
</asy>
 +
- Diagram by Brendanb4321
 +
 +
 +
Extend <math>AB</math> to form a right triangle with legs <math>6</math> and <math>8</math> such that <math>AD</math> is the hypotenuse and connect the points <math>CD</math> so
 +
that you have a rectangle. (We know that <math>\triangle ADE</math> is a <math>6-8-10</math>, since <math>\triangle DEB</math> is an <math>8-15-17</math>.) The base <math>CD</math> of the rectangle will be <math>9+6+6=21</math>. Now, let <math>O</math> be the intersection of <math>BD</math> and <math>AC</math>. This means that <math>\triangle ABO</math> and <math>\triangle DCO</math> are with ratio <math>\frac{21}{9}=\frac73</math>. Set up a proportion, knowing that the two heights add up to 8. We will let <math>y</math> be the height from <math>O</math> to <math>DC</math>, and <math>x</math> be the height of <math>\triangle ABO</math>.
 +
<cmath>\frac{7}{3}=\frac{y}{x}</cmath>
 +
<cmath>\frac{7}{3}=\frac{8-x}{x}</cmath>
 +
<cmath>7x=24-3x</cmath>
 +
<cmath>10x=24</cmath>
 +
<cmath>x=\frac{12}{5}</cmath>
 +
 +
This means that the area is <math>A=\tfrac{1}{2}(9)(\tfrac{12}{5})=\tfrac{54}{5}</math>. This gets us <math>54+5=\boxed{059}.</math>
 +
 +
-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers
 +
 +
==Solution 2==
 +
 +
Using the diagram in Solution 1, let <math>E</math> be the intersection of <math>BD</math> and <math>AC</math>. We can see that angle <math>C</math> is in both
 +
<math>\triangle BCE</math> and <math>\triangle ABC</math>. Since <math>\triangle BCE</math> and <math>\triangle ADE</math> are congruent by AAS, we can then state <math>AE=BE</math> and <math>DE=CE</math>. It follows that <math>BE=AE</math> and <math>CE=17-BE</math>. We can now state that the area of <math>\triangle ABE</math> is the area of <math>\triangle ABC-</math> the area of <math>\triangle BCE</math>. Using Heron's formula, we compute the area of <math>\triangle ABC=36</math>. Using the Law of Cosines on angle <math>C</math>, we obtain
 +
 +
<cmath>9^2=17^2+10^2-2(17)(10)cosC</cmath>
 +
<cmath>-308=-340cosC</cmath>
 +
<cmath>cosC=\frac{308}{340}</cmath>
 +
(For convenience, we're not going to simplify.)
 +
 +
Applying the Law of Cosines on <math>\triangle BCE</math> yields
 +
<cmath>BE^2=10^2+(17-BE)^2-2(10)(17-BE)cosC</cmath>
 +
<cmath>BE^2=389-34BE+BE^2-20(17-BE)(\frac{308}{340})</cmath>
 +
<cmath>0=389-34BE-(340-20BE)(\frac{308}{340})</cmath>
 +
<cmath>0=389-34BE+\frac{308BE}{17}</cmath>
 +
<cmath>0=81-\frac{270BE}{17}</cmath>
 +
<cmath>81=\frac{270BE}{17}</cmath>
 +
<cmath>BE=\frac{51}{10}</cmath>
 +
This means <math>CE=17-BE=17-\frac{51}{10}=\frac{119}{10}</math>. Next, apply Heron's formula to get the area of <math>\triangle BCE</math>, which equals <math>\frac{126}{5}</math> after simplifying. Subtracting the area of <math>\triangle BCE</math> from the area of <math>\triangle ABC</math> yields the area of <math>\triangle ABE</math>, which is <math>\frac{54}{5}</math>, giving us our answer, which is <math>54+5=\boxed{059}.</math>
 +
-Solution by flobszemathguy
 +
 +
==Solution 3 (Very quick)==
 +
<asy>
 +
unitsize(10);
 +
pair A = (0,0);
 +
pair B = (9,0);
 +
pair C = (15,8);
 +
pair D = (-6,8);
 +
draw(A--B--C--cycle);
 +
draw(B--D--A);
 +
label("$A$",A,dir(-120));
 +
label("$B$",B,dir(-60));
 +
label("$C$",C,dir(60));
 +
label("$D$",D,dir(120));
 +
label("$9$",(A+B)/2,dir(-90));
 +
label("$10$",(D+A)/2,dir(-150));
 +
label("$10$",(C+B)/2,dir(-30));
 +
label("$17$",(D+B)/2,dir(60));
 +
label("$17$",(A+C)/2,dir(120));
 +
 +
draw(D--(-6,0)--A,dotted);
 +
label("$8$",(D+(-6,0))/2,dir(180));
 +
label("$6$",(A+(-6,0))/2,dir(-90));
 +
 +
draw((4.5,0)--(4.5,2.4),dotted);
 +
label("$h$", (4.5,1.2), dir(180));
 +
label("$4.5$", (6,0), dir(90));
 +
 +
</asy>
 +
- Diagram by Brendanb4321 extended by Duoquinquagintillion
 +
 +
Begin with the first step of solution 1, seeing <math>AD</math> is the hypotenuse of a <math>6-8-10</math> triangle and calling the intersection of <math>DB</math> and <math>AC</math> point <math>E</math>. Next, notice <math>DB</math> is the hypotenuse of an <math>8-15-17</math> triangle. Drop an altitude from <math>E</math> with length <math>h</math>, so the other leg of the new triangle formed has length <math>4.5</math>. Notice we have formed similar triangles, and we can solve for <math>h</math>.
 +
 +
<cmath>\frac{h}{4.5} = \frac{8}{15}</cmath>
 +
<cmath>h = \frac{36}{15} = \frac{12}{5}</cmath>
 +
 +
So <math>\triangle ABE</math> has area <cmath>\frac{ \frac{12}{5} \cdot 9}{2} = \frac{54}{5}</cmath>
 +
And <math>54+5=\boxed{059}.</math>
 +
- Solution by Duoquinquagintillion
 +
 +
== Solution 4 ==
 +
Let <math>a = \angle{CAB}</math>. By Law of Cosines,
 +
<cmath>\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}</cmath>
 +
<cmath>\sin a = \sqrt{1-\cos^2 a} = \frac{8}{17}</cmath>
 +
<cmath>\tan a = \frac{8}{15}</cmath>
 +
<cmath>A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}</cmath>
 +
And <math>54+5=\boxed{059}.</math>
 +
 +
- by Mathdummy
 +
 +
== Solution 5 ==
 +
Because <math>AD = BC</math> and <math>\angle BAD = \angle ABC</math>, quadrilateral <math>ABCD</math> is cyclic. So, Ptolemy's theorem tells us that
 +
<cmath>AB \cdot CD + BC \cdot AD = AC \cdot BD \implies 9 \cdot CD + 10^2 = 17^2 \implies CD = 21.</cmath>
 +
 +
From here, there are many ways to finish which have been listed above. If we let <math>AB \cap CD = P</math>, then
 +
<cmath>\triangle APB \sim \triangle CPD \implies \frac{AP}{AB} = \frac{CP}{CD} \implies \frac{AP}{9} = \frac{17-AP}{21} \implies AP = 5.1.</cmath>
 +
 +
Using Heron's formula on <math>\triangle ABP</math>, we see that
 +
<cmath>[ABC] = \sqrt{9.6(9.6-5.1)(9.6-5.1)(9.6-9)} = 10.8 = \frac{54}{5}.</cmath>
 +
 +
Thus, our answer is <math>059</math>. ~a.y.711
 +
 +
== Solution 6 ==
 +
 +
Let <math>A=(0,0), B=(9,0)</math>. Now consider <math>C</math>, and if we find the coordinates of <math>C</math>, by symmetry about <math>x=4.5</math>, we can find the coordinates of D.
 +
 +
So let <math>C=(a,b)</math>. So the following equations hold:
 +
 +
<math>\sqrt{(a-9)^2+(b)^2}=17</math>.
 +
 +
<math>\sqrt{a^2+b^2}=10</math>.
 +
 +
Solving by squaring both equations and then subtracting one from the other to eliminate <math>b^2</math>, we get <math>C=(-6,8)</math> because <math>C</math> is in the second quadrant.
 +
 +
Now by symmetry, <math>D=(16, 8)</math>.
 +
 +
So now you can proceed by finding the intersection and then calculating the area directly. We get <math>\boxed{059}</math>.
 +
 +
~hastapasta
 +
 +
== Solution 7 ==
 +
 +
Since the figure formed by connecting the vertices of the congruent triangles is a isoceles trapezoid, by Ptolemys, the other base of the trapezoid is <math>21.</math> Then, dropping altitudes to the base of <math>21</math> and using pythagorean theorem, we have the height is <math>8,</math> and we can use similar triangles to finish.
 +
 +
== Solution 8 (Very, ''very'', quick, but for observant people only) ==
 +
 +
<asy>
 +
//Made by Afly. I used some resources.
 +
//Took me 10 min to get everything right.
 +
import olympiad;
 +
unitsize(18);
 +
pair A = (0,0);
 +
pair B = (0,8);
 +
pair C = (6,0);
 +
pair D = (15,0);
 +
pair E = (21,0);
 +
pair F = (21,8);
 +
pair G = (21/2,0);
 +
pair H = intersectionpoints(B--D,C--F)[0];
 +
pen dash1 = linetype(new real [] {9,9})+linewidth(1);
 +
pen solid1 = linetype(new real [] {9,0})+linewidth(1);
 +
pen dash2 = linetype(new real [] {3,3})+linewidth(1);
 +
fill(C--G--H--cycle,rgb(3/4,1/4,1/4));
 +
fill(D--G--H--cycle,rgb(3/4,3/4,1/4));
 +
draw(C--A--B,dash1);
 +
draw(C--B--D--C,solid1);
 +
draw(F--E--D,dash1);
 +
draw(F--D--C--F,solid1);
 +
draw(G--H,dash2);
 +
draw(brace(D+dir(270),A+dir(270)),solid1);
 +
draw(brace(D,C),solid1);
 +
draw(A--A+2*dir(180),dash1,EndArrow);
 +
draw(E--E+2*dir(0),dash1,EndArrow);
 +
pair L1 = (15/2,-7/2);
 +
pair L2 = (21/2,-13/8);
 +
label("15",L1);
 +
label("8",A--B,W);
 +
label("6",A--C,S);
 +
label("10",B--C,SW);
 +
label("17",B--D,NE);
 +
label("9",L2);
 +
label("4.5",G--D,S);
 +
label("2.4",G--H,W);
 +
markscalefactor = 1/16;
 +
draw(rightanglemark(H,G,D));
 +
draw(rightanglemark(B,A,C));
 +
draw(rightanglemark(D,E,F));
 +
label("A",C,SW);
 +
label("B",D,SE);
 +
label("C",B,NW);
 +
label("D",F,NE);
 +
label("E",A,SW);
 +
label("F",E,SE);
 +
label("G",G,NW);
 +
label("H",H,N);
 +
</asy>
 +
 +
First, let's define H as the intersection of CB and DA, and define G as the midpoint of AB. Next, let E and F be the feet of the perpendicular lines from C and D to line AB respectively. We get a pleasing line of symmetry HG where A maps to B, C maps to D, and E maps to F. We notice that 8-15-17 and 8-6-10 are both pythagorean triples and we test that theory. Then, we find AB = 15 - 6 = 9 and GB = 4.5. Since △CEB~△HGB, we find that HG = 2.4. Then, the area of △HGB is 5.4 and the total overlap is 10.8 = 54/5. Noting that GCD(54,5)=1, we add them to get 59.
 +
 +
Note: I omitted some computation
 +
 +
~ [[User:Afly|Afly]] ([[User talk:Afly|talk]])
 +
 +
==See Also==
 +
{{AIME box|year=2019|n=II|before=First Problem|num-a=2}}
 +
[[Category: Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 13:02, 7 September 2024

Problem

Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

[asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); pair E = (-6,0); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$E$",E,dir(-135)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120)); draw(D--E--A,dotted); label("$8$",(D+E)/2,dir(180)); label("$6$",(A+E)/2,dir(-90)); [/asy] - Diagram by Brendanb4321


Extend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so that you have a rectangle. (We know that $\triangle ADE$ is a $6-8-10$, since $\triangle DEB$ is an $8-15-17$.) The base $CD$ of the rectangle will be $9+6+6=21$. Now, let $O$ be the intersection of $BD$ and $AC$. This means that $\triangle ABO$ and $\triangle DCO$ are with ratio $\frac{21}{9}=\frac73$. Set up a proportion, knowing that the two heights add up to 8. We will let $y$ be the height from $O$ to $DC$, and $x$ be the height of $\triangle ABO$. \[\frac{7}{3}=\frac{y}{x}\] \[\frac{7}{3}=\frac{8-x}{x}\] \[7x=24-3x\] \[10x=24\] \[x=\frac{12}{5}\]

This means that the area is $A=\tfrac{1}{2}(9)(\tfrac{12}{5})=\tfrac{54}{5}$. This gets us $54+5=\boxed{059}.$

-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers

Solution 2

Using the diagram in Solution 1, let $E$ be the intersection of $BD$ and $AC$. We can see that angle $C$ is in both $\triangle BCE$ and $\triangle ABC$. Since $\triangle BCE$ and $\triangle ADE$ are congruent by AAS, we can then state $AE=BE$ and $DE=CE$. It follows that $BE=AE$ and $CE=17-BE$. We can now state that the area of $\triangle ABE$ is the area of $\triangle ABC-$ the area of $\triangle BCE$. Using Heron's formula, we compute the area of $\triangle ABC=36$. Using the Law of Cosines on angle $C$, we obtain

\[9^2=17^2+10^2-2(17)(10)cosC\] \[-308=-340cosC\] \[cosC=\frac{308}{340}\] (For convenience, we're not going to simplify.)

Applying the Law of Cosines on $\triangle BCE$ yields \[BE^2=10^2+(17-BE)^2-2(10)(17-BE)cosC\] \[BE^2=389-34BE+BE^2-20(17-BE)(\frac{308}{340})\] \[0=389-34BE-(340-20BE)(\frac{308}{340})\] \[0=389-34BE+\frac{308BE}{17}\] \[0=81-\frac{270BE}{17}\] \[81=\frac{270BE}{17}\] \[BE=\frac{51}{10}\] This means $CE=17-BE=17-\frac{51}{10}=\frac{119}{10}$. Next, apply Heron's formula to get the area of $\triangle BCE$, which equals $\frac{126}{5}$ after simplifying. Subtracting the area of $\triangle BCE$ from the area of $\triangle ABC$ yields the area of $\triangle ABE$, which is $\frac{54}{5}$, giving us our answer, which is $54+5=\boxed{059}.$ -Solution by flobszemathguy

Solution 3 (Very quick)

[asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120)); draw(D--(-6,0)--A,dotted); label("$8$",(D+(-6,0))/2,dir(180)); label("$6$",(A+(-6,0))/2,dir(-90)); draw((4.5,0)--(4.5,2.4),dotted); label("$h$", (4.5,1.2), dir(180)); label("$4.5$", (6,0), dir(90)); [/asy] - Diagram by Brendanb4321 extended by Duoquinquagintillion

Begin with the first step of solution 1, seeing $AD$ is the hypotenuse of a $6-8-10$ triangle and calling the intersection of $DB$ and $AC$ point $E$. Next, notice $DB$ is the hypotenuse of an $8-15-17$ triangle. Drop an altitude from $E$ with length $h$, so the other leg of the new triangle formed has length $4.5$. Notice we have formed similar triangles, and we can solve for $h$.

\[\frac{h}{4.5} = \frac{8}{15}\] \[h = \frac{36}{15} = \frac{12}{5}\]

So $\triangle ABE$ has area \[\frac{ \frac{12}{5} \cdot 9}{2} = \frac{54}{5}\] And $54+5=\boxed{059}.$ - Solution by Duoquinquagintillion

Solution 4

Let $a = \angle{CAB}$. By Law of Cosines, \[\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}\] \[\sin a = \sqrt{1-\cos^2 a} = \frac{8}{17}\] \[\tan a = \frac{8}{15}\] \[A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}\] And $54+5=\boxed{059}.$

- by Mathdummy

Solution 5

Because $AD = BC$ and $\angle BAD = \angle ABC$, quadrilateral $ABCD$ is cyclic. So, Ptolemy's theorem tells us that \[AB \cdot CD + BC \cdot AD = AC \cdot BD \implies 9 \cdot CD + 10^2 = 17^2 \implies CD = 21.\]

From here, there are many ways to finish which have been listed above. If we let $AB \cap CD = P$, then \[\triangle APB \sim \triangle CPD \implies \frac{AP}{AB} = \frac{CP}{CD} \implies \frac{AP}{9} = \frac{17-AP}{21} \implies AP = 5.1.\]

Using Heron's formula on $\triangle ABP$, we see that \[[ABC] = \sqrt{9.6(9.6-5.1)(9.6-5.1)(9.6-9)} = 10.8 = \frac{54}{5}.\]

Thus, our answer is $059$. ~a.y.711

Solution 6

Let $A=(0,0), B=(9,0)$. Now consider $C$, and if we find the coordinates of $C$, by symmetry about $x=4.5$, we can find the coordinates of D.

So let $C=(a,b)$. So the following equations hold:

$\sqrt{(a-9)^2+(b)^2}=17$.

$\sqrt{a^2+b^2}=10$.

Solving by squaring both equations and then subtracting one from the other to eliminate $b^2$, we get $C=(-6,8)$ because $C$ is in the second quadrant.

Now by symmetry, $D=(16, 8)$.

So now you can proceed by finding the intersection and then calculating the area directly. We get $\boxed{059}$.

~hastapasta

Solution 7

Since the figure formed by connecting the vertices of the congruent triangles is a isoceles trapezoid, by Ptolemys, the other base of the trapezoid is $21.$ Then, dropping altitudes to the base of $21$ and using pythagorean theorem, we have the height is $8,$ and we can use similar triangles to finish.

Solution 8 (Very, very, quick, but for observant people only)

[asy] //Made by Afly. I used some resources. //Took me 10 min to get everything right. import olympiad; unitsize(18); pair A = (0,0); pair B = (0,8); pair C = (6,0); pair D = (15,0); pair E = (21,0); pair F = (21,8); pair G = (21/2,0); pair H = intersectionpoints(B--D,C--F)[0]; pen dash1 = linetype(new real [] {9,9})+linewidth(1); pen solid1 = linetype(new real [] {9,0})+linewidth(1); pen dash2 = linetype(new real [] {3,3})+linewidth(1); fill(C--G--H--cycle,rgb(3/4,1/4,1/4)); fill(D--G--H--cycle,rgb(3/4,3/4,1/4)); draw(C--A--B,dash1); draw(C--B--D--C,solid1); draw(F--E--D,dash1); draw(F--D--C--F,solid1); draw(G--H,dash2); draw(brace(D+dir(270),A+dir(270)),solid1); draw(brace(D,C),solid1); draw(A--A+2*dir(180),dash1,EndArrow); draw(E--E+2*dir(0),dash1,EndArrow); pair L1 = (15/2,-7/2); pair L2 = (21/2,-13/8); label("15",L1); label("8",A--B,W); label("6",A--C,S); label("10",B--C,SW); label("17",B--D,NE); label("9",L2); label("4.5",G--D,S); label("2.4",G--H,W); markscalefactor = 1/16; draw(rightanglemark(H,G,D)); draw(rightanglemark(B,A,C)); draw(rightanglemark(D,E,F)); label("A",C,SW); label("B",D,SE); label("C",B,NW); label("D",F,NE); label("E",A,SW); label("F",E,SE); label("G",G,NW); label("H",H,N); [/asy]

First, let's define H as the intersection of CB and DA, and define G as the midpoint of AB. Next, let E and F be the feet of the perpendicular lines from C and D to line AB respectively. We get a pleasing line of symmetry HG where A maps to B, C maps to D, and E maps to F. We notice that 8-15-17 and 8-6-10 are both pythagorean triples and we test that theory. Then, we find AB = 15 - 6 = 9 and GB = 4.5. Since △CEB~△HGB, we find that HG = 2.4. Then, the area of △HGB is 5.4 and the total overlap is 10.8 = 54/5. Noting that GCD(54,5)=1, we add them to get 59.

Note: I omitted some computation

~ Afly (talk)

See Also

2019 AIME II (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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All AIME Problems and Solutions

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