Math Olympiad | Find the area of the square inside the triangle | 2 Different Methods

[@SALogics]

Very nine explanation

[@stevenjeng1237]

It is easier to let AP=x, then QC=10-2x=5x, so x=10/7, closed

[@prime423]

The most important part is not the mechanics but the plan., Here we have 4 similar triangles plus the fact PQ,a side of the square is parallel to the base of the large triangle. Can we use these observations to solve this problem? The answer is yes with more than one solution.

[@murdock5537]

(10√5/7)^2

[@lasalleman6792]

Or side of square * (10/5 +5/10 +1) = hypotenuse. Here 11.18/3.5 ; side of square: 3.19 Area of square 10.20

[@michaeldoerr5810]

This is the fifth video in which I have learned both methods and have found both methods more useful than the comments. I shall put that problem as geometry practice.

[@jimlocke9320]

We find that ΔABC, ΔABC, ΔAPQ, ΔPBS and ΔCQR are similar. Lengths AB and AC are given and length BC is found by the Pythagorean theorem to be 5√5. The ratio of sides for the similar triangles (short : long : hypotenuse) is 5:10:5√5, which simplifies to 1:2:√5. As in the video, let the length of the side of the square be x. Then, by ratios of sides of similar triangles, AP = x/(√5) and PB = x(√5)/2. However, AB = 5 (given) and AP + PB = AB. So, x/(√5) + x(√5)/2 = 5. Multiply both sides by √5 and simplify: x + 5x/2 = 5√5, 7x/2 = 5√5, x = (10√5)/7. Area of square = x² = ((10√5)/7)² = 500/49, as Math Booster also found.

[@hongningsuen1348]

Alternative method using trigonometric ratio: 1. BC = sqrt 125 (by Pythagoras theorem) 2. Angle ABC = angle APQ (by corresponding angles of // lines BC & PQ, or by similar triangles ABC ~ APQ) Let the angle be theta. 3. From triangle ABC, cos(theta) = 1/sqrt 5 and sin(theta) = 2/sqrt 5. 4. Let side of square PQRS be x. 5. In triangle PBS, PB = x/sin(theta) = (x sqrt 5)/2 6. In triangle APQ, AP = xcos(theta) = x/sqrt 5 7. PB + AP = 5 (given) Hence (x sqrt 5)/2 + x/sqrt 5 = 5 x = (10 sqrt 5)/7 8. Area of square = x^2 = 500/49.

[@quigonkenny]

Edit: Watching the video after my post, at 14:25, you can save a step when comparing ∆ABC and ∆APQ. You don't need to use similar triangles to find a value for AM with respect to x. You already have one, and with it you can directly calculate x: BC/PQ = AN/AM 5√5/x = 2√5/[2√5-x]

[@devondevon4366]

10.209 approximately 10.209 The presence of the square turns all 4 triangles into a similar triangle Hence, all the triangles are similar Notice that one leg is twice the other (5 vs. 10), or one is half the other. Using Pythagorean, the hypotenuse = 11.10834 Let label the square n. label all four sides Hence, the smallest leg of the triangle to the left = 0.5 n and the long leg other triangle to the right = 2n Hence, the hypotenuse of the large triangle in terms of n = 0.5n + 2n + n = 3.5 n But 3.5n = 11.1834, the length of the hypotenuse Hence n= 11.1834/3.5 n= 3.193811 Hence, the area of the square = n^2 or 3.193811^2 = 10.209Answerr

[@devondevon4366]

16:32 500/49 = 10.2