2019 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 15 Medium

Right triangles T_1 and T_2 have areas 1 and 2, respectively. A side of T_1 is congruent to a side of T_2 and a different side of T_1 is congruent to a different side of T_2. What is the square of the product of the other (third) sides of T_1 and T_2? (2019 AMC 10B Problem, Question#15)

  • A.

    \frac{28}{3}

  • B.

    10

  • C.

    \frac{32}{3}

  • D.

    \frac{34}{3}

  • E.

    12

Answer:A

First of all, let the two sides which are congruent be x and y, where y>x. The only way that the conditions of the problem can be satisfied is if x is the shorter leg of T_2 and the longer leg of T_1, and y is the longer leg of T_2 and the hypotenuse of T_1.

Notice that this means the value we are looking for is the square of \sqrt{x^{2}+y^{2}}\cdot \sqrt{y^{2}-x^{2}}= \sqrt{y^{4}-x^{4}}, which is just y^4-x^4.

The area conditions give us two equations: \frac{xy}{2}=2 and \frac{x \sqrt{y^{2}-x^{2}}}{2}=1.

This means that y= \frac{4}{x} and that \frac{4}{x^{2}}=y^{2}-x^{2}.

Taking the second equation, we get x^2y^2-x^4=4, so since xy=4, x^4=12.

Since y= \frac{4}{x}, we got y^4=\frac {256}{12}=\frac {64}{3}.

The value we are looking for is just y^{4}-x^{4}= \frac{64-36}{3}= \frac{28}{3} so the answer \text {(A)}\frac{28}{3}.

Solution by Invoker.

Like in solution 1, we have \frac{xy}{2}=2 and \frac{x \sqrt{y^{2}-x^{2}}}{2}=1.

Squaring both equations yieds x^2y^2=16 and x^2(y^2-x^2)=4,

Let a=x^{2} and b=y^{2}, Then b= \frac{16}{a}, and a(\frac {16}{a}-a)=4\Rightarrow 16-a^2=4\Rightarrow a=2\sqrt 3,

so b= \frac{16}{2 \sqrt{3}}= \frac{8 \sqrt{3}}{3}

We are looking for the value of y^4-x^4=b^2-a^2 ,so the answers is \frac{64}{3}-12=\text {(A)}\frac{28}{3}.

Firstly, let the right triangles be \triangle ABC and \triangle EDF, with \triangle ABC being the smaller triangle. As ir Solution 1, let \overline{AB}= \overline{EF}=x and \overrightarrow{BC}= \overline{DF}=y, Additionally let  \overrightarrow{AC}=z and \overline{DE}=w,

We are given that [ABC]=1 and [EDF]=2, so using area= \frac{bh}{2}, we have  \frac{xy}{2}=1 and \frac{xw}{2}=2. Dviding the two equations, we get \frac{xy}{xw}= \frac{y}{w}=2, so y=2w,

Thus \triangle EDF is a 30^\circ-60^\circ-90^\circ right triangle, meaning that x=w \sqrt{3}. Now by the Pyhagorean Theorem in {\triangle ABC}, (w \sqrt{3})^{2}+(2w)^{2}=z^{2}\Rightarrow 3w^{2}+4w^{2}=z^{2}\Rightarrow 7w^{2}=z^{2}\Rightarrow w \sqrt{7}=z.

The problem requires the square of the product of the third side lengths of each triangle, which is (wz)^2 By substitution, we see that wz=(w)(w \sqrt{7})=w^{2}\sqrt{7}. We also know \frac {xw}{2}=1 \Rightarrow \frac{(w)(w \sqrt{3})}{2}=1 \Rightarrow(w)(w \sqrt{3})=2 \Rightarrow w^{2}\sqrt{3}=2 \Rightarrow w^{2}= \frac{2 \sqrt{3}}{3}. Since we want (w^2\sqrt 7)^2, multiplying both sides by \sqrt 7 get us w^2\sqrt 7=\frac{2 \sqrt{21}}{3}. Now squaring gives (\frac{2 \sqrt{21}}{3})^{2}= \frac{4*21}{9}=\text {(A)}\frac{28}{3}.

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