AIME 2019 II · 第 1 题

Solution

• 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 similar 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)

• 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)

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)

Solution 9 ( Slightly easier than some of the solutions)

Proceed the same as solution one to get the ratio of $\triangle ABO$ to $\triangle DCO$ is $\frac37$. Call BO as x and call DO as $\frac73$*x. Notice how they sum to $17$ as DB=$17$. This means that $\frac73$*x+ x =$17$. Solve the linear equation to get x=$5.1$. Notice how triangle $ABC$ is isosceles with height $\sqrt{5.1^2-4.5^2} or 2.4$. The area is $2.4$*$9$/$2$ or $54/5$. Therefore, the answer is $54+5=\boxed{059}.$

-AD_12