2020 AMC 10B Problems/Problem 8

Revision as of 16:27, 7 February 2020 by Quacker88 (talk | contribs) (See Also)

Solution

There are $3$ options here:

1. $\textbf{P}$ is the right angle.

It's clear that there are $2$ points that fit this, one that's directly to the right of $P$ and one that's directly to the left. We don't need to find the length, we just need to know that it is possible, which it is.

2. $\textbf{Q}$ is the right angle.

Using the exact same reasoning, there are also $2$ solutions for this one.

3. The new point is the right angle.

[asy]  pair  A, B, C, D, X, Y; A = (0,0); B = (0,8); C = (3,6.64575131106); D = (0,6.64575131106); X = (0,4); Y = (1.5,6.64575131106);   draw(A--B--C--A); draw(C--D);  label("$8$", X, W); label("$3$", Y, S);  dot("$A$", A, S); dot("$B$", B, N); dot("$C$", C, E);  draw(rightanglemark(A, C, B)); draw(rightanglemark(A, D, C));  Label AB= Label("$8$", position=MidPoint);  [/asy]

The diagram looks something like this. We know that the altitude to base $\overline{AB}$ must be $3$ since the area is $12$. From here, we must see if there are valid triangles that satisfy the necessary requirements.

First of all, $\frac{\overline{BC}\cdot\overline{AC}}{2}=12 \implies \overline{BC}\cdot\overline{AC}=24$ because of the area.

Next, $\overline{BC}^2+\overline{AC}^2=64$ from the Pythagorean Theorem.

From here, we must look to see if there are valid solutions. There are multiple ways to do this:

$\textbf{Recognizing min \& max:}$

We know that the minimum value of $\overline{BC}^2+\overline{AC}^2=64$ is when $\overline{BC} = \overline{AC} = \sqrt{24}$. In this case, the equation becomes $24+24=48$, which is LESS than $64$.

Another possibility is if $\overline{BC}=1, \overline{AC} =24$. The equation becomes $1+576=577$, which is obviously greater than $64$. We can conclude that there are values for $\overline{BC}$ and $\overline{AC}$ in between that satisfy the Pythagorean Theorem.

And since $\overline{BC} \neq \overline{AC}$, the triangle is not isoceles, meaning we could reflect it over $\overline{AB}$ and/or the line perpendicular to $\overline{AB}$ for a total of $4$ triangles this case.

$2+2+4=\boxed{\textbf{D) }8}$ ~quacker88

$\textbf{Completing the square}$

Set the two values to be $x$ and $y$. We know that $xy=24, x^2+y^2=64$. Adding $2$ times equation $1$ to equation $2$ gives us $x^2+2xy+y^2=108 \implies (x+y)^2=108 \implies x+y=\sqrt{108}$. Plugging $x=\sqrt{108}-y$ into the first equation and rearraging, we get $y^2-y\sqrt{108}+24$. The discriminant is $108-4(24)=12$, which is positive, so there are two solutions. However, we got the two solutions on the right side of $\overline{AB}$, there are two more on the left of $\overline{AB}$ by symmetry. Again, $2+2+4=\boxed{\textbf{D) }8}$

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 2
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 10 Problems and Solutions

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