France Math Olympiad Problem | A Very Nice Geometry Challenge 

At 4:25, Math Booster has used the Pythagorean theorem to determine that BC = 5√5 and AC = 11. Later, the intersection of AC and BD is labelled P. We note that ΔAPB and ΔCDP are similar by angle - angle (

If we let BE = x then side length CE we can calculate from Pythagorean identity ABE is similar to EDC by angle measures equality x/2 = sqrt((10-x)^2+5^2)/5 (x/2)^2 - ((10-x)^2+5^2)/25 = 0 x^2/4 - ((10-x)^2)/25 - 1 = 0 x^2/4 - (x^2-20x+100)/25 - 1 = 0 x^2/4 -x^2/25 +4/5x-5 = 0 21x^2 + 80x - 500 =0 80^2-4*21*(-500) = 6400+42000 80^2-4*21*(-500) =48400 x = (-80+220)/42 x = 140/42 x = 10/3 Area = 1/2*10*5 - 1/2*(10-10/3)*5 Area = 5/2 (10-(10-10/3)) Area = 5/2*10/3 Area = 25/3

*Solução Elegante:* Usando o teorema de Pitágoras no triângulo BCD, temos: (BC)^2=125 e usando novamente o teorema de Pitágoras no triângulo ABC, teremos que: (AC)^2= (BC)^2 - (AD)^2, donde BC=11. Seja P o ponto de interseção dos segmentos AC e BD. claramente os triângulos ABP e CDP são semelhantes, pois os ângulos BAP=CPD=90° e APB=DPC(opostos pelo vértice). Vamos denotar W=[ABP] e H=[DPC]. Pela semelhança de triângulos, temos: W/H=(2/5)^2=4/25, ou seja, *W=4H/25.* Note que: [BDC] - [ABC]= H - W= = 50/2- 22/2 =25 -11=14. Assim, H - 4H/25= 14 => 21H/25=14 => *H=50/3.* Seja S=[BPC]. Assim, S+H=[BDC]=25, logo: S=25 -50/3=> *S=25/3.*

The hypotenuse is 5 SQRT(5) [SQRT(125)]. AC is thus SQRT(125 - 4) = 11. What we really need is the height of the intersection of DB and AC. Let’s call that point I This will be a bit of a challenge. But we should notes that BAI and CAI, despite this content provides weak cartographical skills, are similar triangles. So the linear scale of BAI is 2/5ths that of CDI. Isn’t life grand [D is a right angle near the event horizon of a black hole]. So that Areas: BAC + BDC = Areas: BAI + CDI + 2 BCI = 36 BAC = 11 and BDC = 25 (by inspection) = 36 We already know 1 thing, the area is less than 18, in fact we can kind of guess that DCI is greater than 12.5 ABI is 25/2 * 4/25 = 2 so those combined are ~15 (use 7.5) AIB < (18-7.5) or 10.5 the height is at least 1.25 so 1.25 /2 * 11 =13.75/2 = 6.875 so 6.9 < A < 10.5 rough guess, any answer outside of this range are invalid. OK so here we what is the height of D above BC the height vector creates a similar triangle with ADC since its hypotenuse is 5, the long side on the unit triangle is 2/ SQRT(5) so the height is 5*2/SQRT(5) which is 2 SQRT(5). We need a reference point. Let me choose B. From the height vector BC intercept to C is 1 SQRT(5). Thus from B to the same intercept is 5 SQRT(5) - 1 SQRT(5) = 4 SQRT(5) So if B = (0,0) and C= (5 SQRT(5),0) then D = (4 SQRT(5), 2 SQRT(5)) The slope of BD = 1/2 and it’s y intercept is zero Thus y = x/2 + 0 We have a similar situation with A the height is 2 * 11/5SQRT(5) = 22/ 5SQRT(5) The x coordinate is 2 x 2 / 5SQRT(5) = 4/ 5SQRT(5). The horizontal traverse of AC is 5 SQRT(5)- 4/5 SQRT(5) = (125-4)/ 5SQRT(5) = 121/5SQRT(5) = 21/5 SQRT(5). It’s negative slope is 22/5SQRT(5) / 121/5SQRT(5) = 22/121 = 2/11 Slope is -2/11 0 = SQRT(5) 5*-2 / 11 + y-int. 10 SQRT(5)/11 = y = -2x/11 + 10 SQRT(5)/11 The equation: x/2 = -2x/11 + 10 SQRT(5)/11 11x/22 + 2x/22 = 10 SQRT(5)/11 13x/22 = 10 SQRT(5)/11 13x/2 = 10 SQRT(5) 13x = 20 SQRT(5) x = SQRT(5) 20/13, y = 1/2 x y “height” = SQRT(5) 10/13 preparing to calculate area divide again by 2 : 5 SQRT(5)/13 Area = 5 SQRT(5)/13 * 5 SQRT(5) = SQRT(125)^2 /13 = 125/13= 9.6154 This presenter has the most annoying voice on YT. It’s five cubed five, not “pipe cube pipe”. Hari Krishna only knows what he meant. Despite a common misconception English was a language before India Invented it. Don’t you have a little sister with a sweet voice and better English skills?

El punto "E" es la intersección de BD y AC. La razón de semejanza entre los triángulos CDE y BAE es s=5/2→s²=25/4 =Razón entre sus áreas. BC²=5²+10²=125→ AC=√(125-2²)=11 → Áreas: BAC=2*11/2=11 ; CDB=5*10/2=25. Si AE=a→ Áreas: BAE=2a/2=a→BCE=(11-a) y CDE=a*s²=25a/4 → BAC+CDB=11+25=36 =a+(25a/4)+2(11-a)→ a=56/21→ Área sombreada =11-a =11-(56/21) =25/3 ud². Gracias y un saludo cordial.

Figure ABCD is a cyclic quadrilateral since angles at A and D are right angles. AC and BD are chords let’s say cutting at E . Let BE =a so ED= 10-a and let AE=b so EC = 11-b. We know that |AC|=11 from using pythag. On triangle ABC. Since triangles AEB and DEC are similar 5/2= (11-b)/a ie 5a+2b=22 Since AC and BD are intersecting chords in the circle around the points ABCD it follows that AE mult.byEC= BE mult by ED. So b(11-b)=a(10-a). So 11b-b^2=10a-a^2. Into that equation substitute for a= (22-2b)/5. I ended up with 3(b^2)-41b+88=0. So b=11 and b=8/3. But b cannot =11 so b= 8/3. Area of triangle ABC = 11 units squared and area of triangle ABE= 8/3 units squared so area of pink triangle= 33/3-8/3=25/3.

From Pythagoras, _|BC| = 5√5_ and _|AC| = 11._ _|AP| / 2 = |DP| / 5_ (similar triangles _ABP, CDP)._ ⇒ _|AP| = ²/₅(10 - |BP|)_ ⇒ _|BP| = 10 - ⁵/₂|AP|_ ... ① But also _|BP|² = |AP|² + 4_ ... ② Solve ① and ② to get _|AP| = ⁸/₃, |BP| = ¹⁰/₃_ ∴ _|PC| = 11 - ⁸/₃ = ²⁵/₃_ Now knowing all 3 sides of _ΔPBC_ use Heron's formula to get its area. _s = ½(¹⁰/₃ + ²⁵/₃ + 5√5) = ⅙(35 + 15√5)_ Area _ΔPBC = √[s(s-|BP|)(s - |PC|)(s - |BC|)]_ = _√[(⅙(35 + 15√5)⅙(15 + 15√5)⅙(-15 + 15√5)⅙(35 - 15√5)]_ = _²⁵/₃₆√[(7 + 3√5)(7 - 3√5)(3 + 3√5)(-3 + 3√5)]_ = _²⁵/₃₆√[(49 - 45)(45 - 9)]_ = *_²⁵/₃_*

Solution in exam answer format. 1. Triangles ABC and BCD are right-angled triangles as given. By Pythagoras theorem: BC^2 = BD^2 + CD^2 = 10^2 + 5^2 = 125 AC^2 = BC^2 - AB^2 = 125 - 4 = 121 Hence AC = 11 2. Let intersecting point of AC and BD be E. ∆ABE ~ ∆DCE (AAA) Hence DC/AB = 5/2 = CE/BE = DE/AE (corresponding sides proportional) Let CE be a, BE be b, DE be c, AE be d. Hence a/b = 5/2 (1) c/d = 5/2 (2) AC = AE + CE = a + d = 11 (from above) (3) BD = BE + DE = b + c = 10 (given) (4) 3. Find CE = a for step 4. From (1) and (2), b = (2/5)a and c = (5/2)d Substitution into (4), b + c = (2/3)a + (5/2)d = 10 (5) From (3), d = 11 - a Substitution into (5), (2/5)a + (5/2)(11 - a) = 10 4a + 25(11 - a) = 100 275 - 100 = 25a - 4a 21a = 175 a = 25/3 4. ∆ABC and ∆EBC are equal height triangles. Hence area of ∆EBC:area of ∆ABC = CE:AC = a:11 = 25/3:11 Area of ∆ABC = ½(2)(11) = 11 Hence purple area of ∆EBC = (11)(25/3)/(11) = 25/3

I thought that DP has a lenght about 10 units . Now I got for the area of triangle BCP 20 Square units .

I'll throw out my answer and then come back and itemize the steps: 25/3 = 8-1/3 = 8.33333...

Bruh i foubd 5root5 because i thought pd was 10

Answer 8.3333 My method The triangle above the shaded triangle is a 3-4-5 triangle ( I didn't show how I got it here. But the three sides of the two triangles are easily determined by Pythagorean) It is a 3-4-5 because the angle in triangle BDC = 63.435 degrees, and the angle in triangle. ACB is 10.305 degrees. The difference is 53.13 degrees ( 63.435 - 10.305), and in a 3-4-5 triangle, the angles are 53.13, 36.87, and 90 degrees Hence, the sides are 5, 6.667, and 8.333 (CP = 8.3333) Thus, it is a 3-4-5 triangle scaled up by 1.66666 Let's label the point where the two unshaded triangles intersect P, then DP = 6.6666; hence, the area of the unshaded triangle to the right = 6.6666 * 5/2 = 16.66665 Length of BC = 11.18 (Pythagorean)' Hence length of AC = 11 (Pythagorean Since AC = 11, and CP=8.333, then AP= 2.6666 ( 11-8.3333), The area of APB = 2* 2.6666/2 = 2.6666 Hence, the area of the unshaded region is 19.3332665 (16.6665 + 2.6666) The area of the entire figure is the area of BCD = 25 ( 5* 10*1/2) + the area of APB 2.6666 = 27.6666 So, the area of the entire figure = 27.6666 the area of the shaded = 27.6666 - 19.3332665= 8.3333 Answer