In the diagram to the right, triangle PQR has a right angle

[Luci]

This is a really tough problem impossible to do in two minutes watching the explanation given by MGMAT experts.
I thought I had an easier aproach but at the end it didnt work. Can anyone shed some light in my reasoning to see where is it wrong?




Remembering that a height to the hypotenuse always divides a right triangle into two smaller triangles that are similar to the original one (since they all have a right angle and they share another of the included angles), therefore all three triangles are similar to each other. Therefore their areas will be in the ratio of the square of their respective side lengths.

We also know that QS has a length of 12.

There are some especial triangles (I think this is where I might be wrong but I would like the experts to confirm it). The right triangles which length sides 3,4,5 and 5,12,13. So using this second especial right triangle if QS is 12 then SR must be 5 and QR 13. And because all the 3 triangles are similar PS 144/5 and PQ 156/5.

Now knowing all the sides comparing areas is easy. The ratio of the area of PQS to the area of QRS equals
(1/2*12*144/5)/(1/2*5*12) but this equals 144/10 which is not one of the solutions given...


I thought those especial right triangles existed, they are consistent with the Pitagoras Theorem. What do you think?

Thanks

[shaji]

These aren't very deep waters.

The key to answering this question within 2 minutes is to realize that the 'rephrased question' is 'the ratio by which the hpotenuse PR is divided by the perpendicular QS?'

Please also note that that the perimeter of PQR 60 , which implies that PQ=20; QR=15 & PR=25.(ESPECIALITY) Now go for the answer choices and certainly 16/9 is the required ratio.

This approach will take hardly a minute.

This is a really tough problem impossible to do in two minutes watching the explanation given by MGMAT experts.
I thought I had an easier aproach but at the end it didnt work. Can anyone shed some light in my reasoning to see where is it wrong?




Remembering that a height to the hypotenuse always divides a right triangle into two smaller triangles that are similar to the original one (since they all have a right angle and they share another of the included angles), therefore all three triangles are similar to each other. Therefore their areas will be in the ratio of the square of their respective side lengths.

We also know that QS has a length of 12.

There are some especial triangles (I think this is where I might be wrong but I would like the experts to confirm it). The right triangles which length sides 3,4,5 and 5,12,13. So using this second especial right triangle if QS is 12 then SR must be 5 and QR 13. And because all the 3 triangles are similar PS 144/5 and PQ 156/5.

Now knowing all the sides comparing areas is easy. The ratio of the area of PQS to the area of QRS equals
(1/2*12*144/5)/(1/2*5*12) but this equals 144/10 which is not one of the solutions given...


I thought those especial right triangles existed, they are consistent with the Pitagoras Theorem. What do you think?

Thanks

[GMAT 2007]

Shaji,

I was able to calculate them by using Area, pythagoras and perimeter of the triangle. But it took me more than a minute for sure.

How did you calculate PQ, QR and RP from the perimeter within a minute? What is ESPECIALITY? Can you shed some more light on this approach?

GMAT 2007

[christiancryan]

This is a *really* tough problem, we admit!

I personally think that the trick to this particular problem is to switch gears at a certain point -- after you write down the equations describing all the relationships, you might realize (as I myself did on first trying this problem) that it would take a long time to derive the solution using algebra. So then you need to say, let me try some common right triangles. 3-4-5 becomes the first candidate -- especially since the perpendicular is 12, which is divisible by both 3 and 4. So a side of 12 (in the two smaller triangles) is easy to "scale" to -- if it's the shortest side (the "3" side), then the longer sides are 16 and 20; likewise, if it's the middle side, the other two sides are 9 and 15. Magically this fits the perimeter constraint (the big triangle's perimeter=60), and you're done.

The 5-12-13 triangle doesn't fit as nicely, and in the end it doesn't yield a big-triangle perimeter of 60 (it actually yields a perimeter of 78 -- try it, making sure that every triangle is similar to a 5-12-13 -- and that the smallest triangle in the figure is actually the 5-12-13 triangle).

[Guest]

Christian has stated this 'ESPECIALITY' for you; ie. The rt triangle 3,4,5 has a perimeter of 12 which is 1/5 the perimeter of 60(triangle PQR). This is realized in less than 15 seconds.

Guest!!! U are missing the important part, the 'rephrase' of the question that I stated. The ratio the hypo of PQR is divided by QS the perpendicular is what is asked by the question setter. Now U know the "rest of the story".

Shaji,

I was able to calculate them by using Area, pythagoras and perimeter of the triangle. But it took me more than a minute for sure.

How did you calculate PQ, QR and RP from the perimeter within a minute? What is ESPECIALITY? Can you shed some more light on this approach?

GMAT 2007

[StaceyKoprince]

Let's simmer down a little guys, okay? We're all here to learn and help each other. :)

I like Chris's approach and I just want to remind you guys - sometimes we're going to hit things that we will never figure out in 2 minutes if we haven't seen something like it before. That happens to EVERYbody on the test - it happens a lot for most people and, even for me, I usually see one or two question set-ups that are completely unfamiliar to me. So, sometimes, we have to just think - what could work here? What can I figure out just from logic / looking at it, or what can I try? (Especially for geometry!)